By Way of Ignorance: Ore Localization

Prologue~
" In order to arrive there,
To arrive where you are, to get from where you are not,
You must go by a way wherein there is no ecstasy.
In order to arrive at what you do not know 
You must go by a way which is the way of ignorance.
In order to possess what you do not possess
You must go by the way of dispossession.
In order to arrive at what you are not
You must go through the way in which you are not."
— T. S. Eliot, Four Quartets (Part II: East Coker)

In the last post we dealt with the concept of localization, categorically, this post, as promised earlier, delves into the Ore condition which helps us extend the concept of localization to non-commutative rings. 

Localization of commutative rings relies heavily on commutativity, but for non-commutative rings multiplication does to commute, and the familiar construction breaks down. The Ore condition provides precisely the compatibility needed to recover a workable theory of fractions in this case. 

Verse I: We begin, again~
To begin with our discussion of the concept of localization for non-commutative rings (NCRs), we rely on the Ore condition.
Let us begin our construction by considering a non-commutative ring $R$ and a multiplicative subset $S$. We seek a ring $S^{-1}R$ and a homomorphism $\iota :R \to S^{-1}R$ such that:

1. $\iota(s)$ is a unit for all $s \in S$,
2. every element of $S^{-1}R$ is of the form $a s^{-1}$ (right fractions) or $s^{-1} a$ (left fractions), and
3. the universal property holds.

Verse II: Some terminologies~

Now, a multiplicative subset $S \subseteq R$ satisfies the right Ore condition if for all $r \in R$ and $s \in S$, the intersection $rS \cap sR$ is nonempty: $$\forall r \in R,\, \forall s \in S,\, \exists\, r' \in R,\, s' \in S \text{ such that } rs' = sr' $$ This condition guarantees the existence of common right multiples, allowing right fractions to be represented over a common denominator. 

Given $a s^{-1}$ and $b t^{-1}$, we can find $s' \in S$, $t' \in R$ such that $s t' = t s'$, allowing us to write: $$a s^{-1} + b t^{-1} = (a t' + b s') (s t')^{-1}$$ $S$ is right reversible if for all $r \in R$, $s \in S$:$$rs = 0 \implies \exists\, t \in S \text{ such that } tr = 0 $$

This ensures the equivalence relation on fractions is well-defined and that $\iota:R \to S^{-1}R$ is injective when $S$ consists of regular elements (non-left and non-right zero divisors).

Verse III: Certain assumptions~
Let us assume that $S$ satisfies the right Ore condition. We define the set of right fractions:$$S^{-1}R = \left\{ (a,s) \mid a \in R,\, s \in S \right\} / \sim $$ where $(a,s) \sim (b,t)$ iff $\exists\, u,v \in R$ with $su = tv \in S$ such that $au = bv$. 

Equivalence relation is then expressed as $\frac{a}{s} = \frac{b}{t} \iff \exists\, u \in R,\, v \in R$ such that $su = tv $ and $au = bv$
Thus, $$S^{-1} R = \{ (a,s) \mid a \in R, s \in S\} / \sim.$$

The Ore condition here, substitues for the lack of commutativity in our ring and defines localization on common, 'agreeable' grounds.

Indeed, given two fractions $\frac{a}{s}, \frac{b}{t},$ the Ore condition provides us with elements $u,v \in R$ such that $su=tv \in S.$ 
Both fractions may therefore be expressed over the common denominator $su=tv,$ and we define their sum by $$\frac{a}{s} + \frac{b}{t} = \frac{a u + b v}{su}.$$

Multiplication gets a bit tricky under these circumstances, since the denominator $s$ simply cannot be moved past the numerator $b$. But (again!) using the Ore condition, we have the existence of a $b' \in R$ and $s' \in S$ such that $sb'=bs'$, and this enables us to define the product as $$\frac{a}{s} \cdot \frac{b}{t} = \frac{ab'}{ts'}$$Symmetrically, $S$ satisfies the left Ore condition if: $$\forall r \in R,\, \forall s \in S,\, \exists\, r' \in R,\, s' \in S \text{ such that } s'r = r's $$

This yields a left ring of fractions $RS^{-1}$ with elements of the form $s^{-1}a$.

If $S$ satisfies both left and right Ore conditions, together with the corresponding reversibility conditions, then the left and right constructions agree canonically, $S^{-1}R \cong RS^{-1},$ giving rise to a two-sided localization.

Verse IV: We have arrived~
The Ore localization is the universal solution to inverting $S$ in the category of (not necessarily commutative) rings:

1. For any $f \colon R \to T$ with $f(S)$ invertible, there exists a unique $\tilde{f} \colon S^{-1}R \to T$ with $\tilde{f} \circ \iota = f$.
2. More generally, for a category $\mathcal{C}$ and class of morphisms $\mathcal{Mor}$, the localized category $\mathcal{C}[\mathcal{Mor}^{-1}]$ exists when $\mathcal{Mor}$ satisfies the Ore conditions (calculus of fractions).
   That is to say: the ring-theoretic construction is a special case of a much broader categorical phenomenon. 

If $R$ is commutative, the Ore condition is guaranteed: $rs = sr$ for all $r \in R$, $s \in S$. Thus every multiplicative subset admits localization.

We are essentially saying that given $R$ to be a ring and $S \subseteq R$ to be a multiplicative subset. The following are equivalent:

1. $S$ satisfies the right Ore condition and right reversible condition.
2. There exists a ring $S^{-1}R$ (the right ring of fractions) and homomorphism $\iota :R \to S^{-1}R$ such that: $\iota(s)$ is a unit for all $s \in S$, every element of $S^{-1}R$ is of the form $a s^{-1}$ with $a \in R$, $s \in S$, $\iota$ satisfies the universal property for inverting $S$.

Epilogue~
" And what you do not know is the only thing you know
And what you own is what you do not own
And where you are is where you are not."
— T. S. Eliot, Four Quartets (Part II: East Coker)

Localization, Categorically

Prologue~
"We shall not cease from exploration
And the end of all our exploring
Will be to arrive where we started
And know the place for the first time."
— T. S. Eliot, Four Quartets

In the previous post, we approached localization through its universal property. Given a ring $R$ and a multiplicative subset $S$, we sought a ring $S^{-1}R$  together with a morphism $ \iota: R \longrightarrow S^{-1}R$  such that every element of $S$ becomes invertible, and such that $S^{-1}R$ is universal with respect to this property.

This point of view is extremely powerful: rather than constructing an object explicitly and then studying its properties, we first describe what the object ought to do and only later worry about how it is built. In many ways, this is one of the recurring themes of modern mathematics. There is, however, another perspective on localization that is equally illuminating. 

Verse I: The Architecture~
The answer is surprisingly elegant: localization can be realized as a colimit. Colimits are usually introduced as a way of gluing together diagrams of objects. Localization, on the other hand, appears to be a process of adjoining inverses. The common link between the two is--when we form the localization $S^{-1}R$, every element $s\in S$ is required to become invertible. Informally, this means that multiplication by $s$, $R\xrightarrow{\cdot s}R$, should become an isomorphism after localization.

Rather than forcing all these inverses to exist simultaneously, we examine the collection of maps induced by the elements of $S$. The multiplicative subset $S$ naturally carries a preorder. Given $s,t\in S$, we write $ s\leq t $ if there exists some $u\in S$ such that $ t=su. $This relation reflects the idea that $t$ is a refinement of the denominator $s$. Since $S$ is multiplicatively closed, any two elements $s$ and $t$ admit a common upper bound, namely $st$. Thus $S$ becomes a filtered preorder.

So essentially, to each $s\in S$, we associate a copy of the $R$-module $R$. Whenever $t=su$, we obtain a morphism $R\xrightarrow{\cdot u}R$. In this way, $S$ determines a filtered diagram of $R$-modules. The remarkable fact is that the localization is precisely the colimit of this diagram. $$ S^{-1}R \cong \varinjlim_{s\in S} R$$

Verse II: A Necessary Interlude~
Before proving the above theorem, let us recall the notion of a filtered diagram and its colimit. Roughly speaking, a filtered diagram is a directed system of objects in which any finite collection of objects can be compared at some later stage. The colimit of such a diagram may be thought of as the object obtained by identifying elements that eventually become equal.

Given a poset $(P, \leq)$, we construct a category whose objects are elements of $P$ and morphisms are $\operatorname{Hom}(x,y)= \{\ast\}$ when $  x\le y$ and $\operatorname{Hom}(x,y)= \varnothing$ otherwise. In this setting, the notion of a filtered category becomes particularly transparent. 

A poset $P$ is said to be filtered if it is nonempty and every pair of elements admits a common upper bound. That is, for any $x,y \in P$, there exists some $z \in P$ such that $x \leq z$ and $y \leq z$. Moreover, since $S$ is multiplicatively closed, any two elements $s,t \in S$ possess a common upper bound, namely $st$, because $s \leq st$ and $t \leq st$. Thus $S$ forms a filtered preorder and naturally serves as an indexing category for the diagram whose colimit will produce the localization $S^{-1}R$.

Verse III: The Limit of the Journey~
A filtered category is very useful because it provides an indexing system for directed collections of objects and morphisms. Let $\mathscr{C}$ be a category and let $I$ be a filtered category. A diagram indexed by $I$ is simply a functor $D: I \to \mathscr{C}$. For each object $i \in I$, the functor assigns an object $D(i) \in \mathscr{C}$, and for each morphism $i \to j$, in $I$, it assigns a morphism $D(i) \to D(j)$ in $\mathscr{C}$. When $I$ is a filtered preorder, we can visualise this data as a directed system $D(i_1)\longrightarrow D(i_2)\longrightarrow D(i_3)\longrightarrow\cdots$ in which any finite collection of objects eventually maps into a common stage.  The colimit of such a diagram is an object that records all the information contained in the system while identifying elements that become equal at sufficiently large stages.

More precisely, let $D:I\to R-\mathbf{Mod} $ be a filtered diagram of $R$-modules. The colimit of $D$, denoted $\varinjlim_{i\in I}D(i), $ is characterized by the following universal property: for every $R$-module $M$, giving a morphism, $\varinjlim_{i\in I}D(i)\to M$ is equivalent to giving a compatible family of morphisms $D(i) \to M$ for all $i \in I$. 
Concretely, filtered colimits of modules admit a particularly simple description. Consider the direct sum $ \bigoplus_{i\in I}D(i). $ Whenever $ f_{ij}:D(i)\to D(j) $ is a morphism in the diagram and $x \in D(i)$, we impose the relation $ x\sim f_{ij}(x)$. The colimit is obtained by quotienting all such relations. $$\varinjlim_{i\in I}D(i) = \left(\bigoplus_{i\in I}D(i)\right)\Big/ \left\langle x-f_{ij}(x) \right\rangle$$

Verse IV: The Familiar Object ~
Now, keeping in mind the above discussions, we delve into the concept of localization. The multiplicative subset $S$ determines a filtered category, and to each $s \in S$ we associate a copy of the $R$-module $R$. The resulting filtered diagram will have colimit $ \varinjlim_{s\in S}R,$ and our goal is to show that this colimit is naturally isomorphic to the localization $S^{-1}R.$

We return to the multiplicative subset $\mathrm{\$S}\backslash subseteqR$. Recall that $S$ carries a filtered preorder given by $ s\leq t \iff t=su$ for some $u\in S.$ To each object $s \in S$, assign the $R$-module $R$. Thus, $ D(s)=R$ whenever $s\leq t, $ say $t=su, $ we define the corresponding morphism $D(s)\longrightarrow D(t) $ to be multiplication by $u: r \longmapsto ru. $

This assignment defines a functor $D:S\to R\text{-}\mathbf{Mod}.$ The resulting diagram therefore has the form $R \xrightarrow{\cdot u} R \xrightarrow{\cdot v} R \xrightarrow{\cdot w} \cdots.$ The colimit $\varinjlim_{s\in S}R$ is obtained by identifying an element $r \in D(s)$ with its image $ru\in D(su).$ In other words, $(r,s)\sim (ru,su)$ for every $r\in R,\quad u\in S.$ And this is precisely the relation which appears in the construction of fractions.

Verse V: The Identification~
The discussion above culminates in the following theorem.
Theorem. Let $R$ be a commutative ring and $S$ a multiplicative subset of $R$. Consider the filtered diagram $D:S\to R\text{-}\mathbf{Mod}$ defined by $D(s)=R$ and $D(s\le t)=\cdot u:R\to R$ whenever $t=su.$. Then localization of $R$ at $S$ is naturally isomorphic to the colimit of this diagram. That is, $S^{-1}R \cong \varinjlim_{s\in S}R. $

Essentially, localization may be realized as the filtered colimit of copies of $R$ connected by multiplication maps coming from the elements of $S$.

Proof. Recall that the colimit of a diagram of $R$-modules may be described as the quotient $\left(\bigoplus_{s\in S}R\right)\Big/\sim,$ where the equivalence relation is generated by $(r,s)\sim (ru,su)$ for every $r\in R,\qquad u\in S.$ 
We define a map $\Phi:\varinjlim_{s\in S}R\longrightarrow S^{-1}R$ by $(r,s) \mapsto \frac{r}{s}$.
We first check that $\Phi$ is well-defined. Suppose $(r,s) \sim (ru,su).$ Then $\Phi(ru,su)=\frac{ru}{su}.$ Since $R$ is commutative, $\frac{ru}{su}=\frac{r}{s},$ and therefore $ \Phi(r,s)=\Phi(ru,su).$

To show surjectivity, observe that every element of $S^{-1}R$ is represented by a fraction $\frac{r}{s}$, which is the image of the class of $(r,s)$.
for injectivity, suppose $ \Phi(r,s)=\Phi(r',t).$. Then $\frac{r}{s}=\frac{r'}{t}$ in $S^{-1}R$. By the definition of localization, there exists $u \in S$ such that $ u(rt-r's)=0.$
Multiplying both pairs by suitable elements of $S$, both $(r,s)$ and $(r',t)$ map to the stage indexed by $stu$. Their images there are $rtu$ and  $r'su,$ which are equal by the previous relation. Hence the two classes coincide in the colimit.

Therefore, $\Phi$ is bijective. Since both constructions carry natural $R$-module structures and $\Phi$ preserves them, $\Phi$ is an isomorphism. Consequently, $S^{-1}R \cong \varinjlim_{s\in S}R.$ 

Epilogue~
"What we call the beginning is often the end
And to make an end is to make a beginning."
— T. S. Eliot, Four Quartets

This post was inspired by the categorical description of localization mentioned in The Rising Sea by Ravi Vakil. In the next post, we will look at the Ore condition and localization for non-commutative rings (NCRs).

The Craft of Localization

Prologue~
“A man sets himself the task of drawing the world.”
— Jorge Luis Borges

In this post, we discuss the concept of localization, a term which originated in algebraic geometry. For a discussion on the term `localization', we refer to the following passage which comes from Eisenbud's "Commutative Algebra with a View Towards Algebraic Geometry."

... localization as a general procedure was defined rather late: In the case of integral domains it was described by Grell, a student of Noether, in 1927, and it was not defined for arbitrary commutative rings until the work of Chevalley [1944] and Uzkov [1948], long after the basic ideas of commutative algebra were established. Perhaps this is because interest was focused on finitely generated algebras on the one hand, and power series rings on the other, and neither of these classes of rings are closed under localization. Instead of passing to a localized rings, as we would now, people often used ideal quotients as a substitute.

Verse I: Surveying the Terrain~
Localization as a philosophy refers to emphasizing on a specific part of a bigger object. So we essentially look closer on a smaller area of it while we tend to, for the time being, ignore the rest. We will discuss the algebraic view of localization, specifically in the context of rings.
Localization of a commutative ring by a multiplicative set is a new ring whose elements are fractions with numerators coming from the ring and the denominators coming from the multipicative set (well,  there's a bit more to it).
We begin by defining a multiplicative set, also known as a multiplicatively closed set, say $S$, which, as the name suggests, consists of elements of the ring, say $R$, such that $S$ forms a subset of $R$ that is closed under multipication and contains the multiplicative identity. The set $S^{-1} R$ will be the resulting localization of $R$ at $S$. 

Verse II: Drawing the Map~
More precisely, a subset $S \subseteq R$ is called a multiplicative set if the following conditions hold:

1. $1 \in S$,
2. $0 \notin S$,
3. for all $a,b \in S$, we have $ab \in S$ (multiplicatively closed).

Then, let \(R\) be a commutative ring with unity and let \(S \subseteq R\) be a multiplicative set. The localization of \(R\) at \(S\) is the ring $S^{-1}R$ consisting of equivalence classes of formal fractions $$ \frac{r}{s}, \quad r\in R,\ s\in S,  \hspace{0.2in} \text{ where }  \hspace{0.2in}\frac{r}{s} \sim \frac{r'}{s'} $$ if there exists \(u\in S\) such that $u(s'r-r's)=0$.
Addition and multiplication are defined by $$\frac{r}{s}+\frac{r'}{s'}  =  \frac{s' r+r's}{ss'}  \hspace{0.2in}\text{  and  } \hspace{0.2in} \frac{r}{s}\cdot\frac{r'}{s'}=\frac{rr'}{ss'}$$ There is a natural ring homomorphism $$ \varphi:R\to S^{-1}R  \hspace{0.2in} \text{given by} \hspace{0.2in} \varphi(r)=\frac{r}{1}$$

Verse III: A different view~

Another way to look at localization is by defining it via the universal property (i.e., in the categorical terms).

Let $R$ be a commutative ring and $S \subseteq R$ a multiplicative subset. The localization $S^{-1}R$ is characterized categorically by a universal property in the category of commutative rings.
The localization consists of:

1. A ring $S^{-1}R$
2. A canonical homomorphism $\iota_S \colon R \to S^{-1}R$

satisfying:

1. Inversion: For every $s \in S$, $\iota_S(s)$ is a unit in $S^{-1}R$.
2. Universality: For any ring homomorphism $f \colon R \to T$ such that $f(s)$ is a unit in $T$ for all $s \in S$, there exists a unique homomorphism $\tilde{f} \colon S^{-1}R \to T$ making the diagram commute:
   
   

This universal property says that $S^{-1}R$ is the initial object in the comma category $(R \downarrow \mathcal{U}_S)$, where $\mathcal{U}_S$ is the full subcategory of the category of commutative rings (say $\mathbf{CRings}$), modulo $R$, i.e., $\mathbf{CRings}/ R$ consisting of morphisms $R \to T$ that invert $S$.

Verse IV: Another (different) view~

Equivalently, localization is the left adjoint to the inclusion functor:

$$L_S \colon \mathbf{CRings} \longrightarrow \mathbf{CRings}_S \quad \text{and} \hspace{0.2in} R \mapsto S^{-1}R$$

where $\mathbf{CRings}_S$ is the category of commutative rings equipped with a homomorphism from $R$ that inverts $S$. The adjunction is: $$ \operatorname{Hom}_{\mathbf{CRings}}(S^{-1}R, T) \cong \operatorname{Hom}_{\mathbf{CRings},\,S^{-1}}(R, T)$$ where the RHS denotes homomorphisms inverting $S$. 

Any two objects satisfying the universal property are uniquely isomorphic via a unique isomorphism commuting with the structure maps.
Localization is functorial in both $R$ and $S$. A ring homomorphism $\varphi \colon R \to R'$ with $\varphi(S) \subseteq S'$ induces $\varphi_S \colon S^{-1}R \to (S')^{-1}R'$.

If $S \subseteq T \subseteq R$, then $(T^{-1}R) \cong T^{-1}(S^{-1}R)$. This follows from the universal property by checking that both sides invert $T$. 

As a set: $S^{-1}R = \left\{ \frac{r}{s} \mid r \in R, \, s \in S \right\} / \sim $ where $\frac{r}{s} \sim \frac{r'}{s'}$  iff  $\exists \ t \in S$ such that $t(rs' - r's) = 0$. Operations are:

$$\frac{r}{s} + \frac{r'}{s'} = \frac{rs' + r's}{ss'}, \quad \frac{r}{s} \cdot \frac{r'}{s'} = \frac{rr'}{ss'}$$

The map $\iota_S(r) = \frac{r}{1}$ inverts $S$ since $\iota_S(s) \cdot \frac{1}{s} = 1$.

Verse V: Toward Ore Localization~

In $\mathbf{CRings}$, every multiplicative subset $S \subseteq R$ admits a localization $S^{-1}R$ because commutativity ensures the fraction calculus works smoothly.
For noncommutative rings, arbitrary multiplicative subsets $S$ may not admit a well-behaved localization. The Ore condition is the necessary and sufficient condition for constructing a ring of fractions $S^{-1}R$ (or $RS^{-1}$) when $R$ is noncommutative. We will be discussing this in depth in the upcomign post.

Epilogue~
“A man sets himself the task of drawing the world. As the years pass, he populates a space with images of provinces, kingdoms, mountains, bays, ships, islands, fishes, rooms, instruments, stars, horses, and people. Shortly before he dies, he discovers that that patient labyrinth of lines traces the image of his own face.”
— Jorge Luis Borges

Beyond Monoids: The Universal Additive Invariant

Prologue~
"He saw that the water continually flowed and flowed and yet it was always there; it was always the same and yet every moment it was new."
—Hermann Hesse, Siddhartha

In mathematics, classification problems often deal with the concept of an invariant. Invariants are primary (and powerful) tools for distinguishing between mathematical objects by contraposition. In this post, we will formalize how the Grothendieck construction universally transforms additive data (abelian monoids) into a group that supports formal subtraction, and how this fundamental algebraic mechanism gives rise to the $K_0$ group of a ring.

Verse I: The ground~
Let $X$ be a set of mathematical objects, and let $\sim$ be an equivalence relation on $X$ (most commonly, isomorphism). An invariant is a well-defined mapping $I: X \to Y$, where $Y$ is some target set of values, such that if $x \sim y$, then $I(x) = I(y)$.
If $I(x) \neq I(y)$, we can definitively state that $x \not\sim y$. However, we often require invariants that do more than merely label objects, since we need them to respect the algebraic structure of $X$.

Suppose $X$ is equipped with a commutative, associative operation, such as the direct sum ($\oplus$). We then seek an additive invariant, which is a mapping into an abelian group $A$ such that $I(x \oplus y) = I(x) + I(y)$. 
While this defines an additive invariant, it leaves a deeper question- does there exist a 'perfect' one? A universal additive invariant through which all others must factor?

Verse II: Abelian Monoids and Group Completion~
A set $M$ equipped with an associative and commutative binary operation ($+$) and an additive identity element ($0$), is called an abelian monoid.
A canonical example is the set of natural numbers, $\mathbb{N}$ (including $0$). While $\mathbb{N}$ is closed under addition, it lacks additive inverses. To construct an invariant that takes values in a group (which mathematically behaves much better than a monoid), we must formally adjoin inverses. This process is known as group completion.

The group completion of an abelian monoid $M$ is an abelian group $M^{-1} M$ equipped with a monoid homomorphism $[\cdot] : M \to M^{-1} M$. This map is universal: for any abelian group $A$ and any monoid homomorphism $\alpha : M \to A$, there exists a unique group homomorphism $\tilde{\alpha} : M^{-1} M \to A$ such that $\tilde{\alpha}([m]) = \alpha(m)$ for all $m \in M$.

Explicitly, we construct the Grothendieck group of $M$, denoted $K_0(M)$, by taking the free abelian group $F(M)$ generated by symbols $[m]$ for all $m \in M$, and factoring out the subgroup $R(M)$ generated by the relations $[m+n] - [m] - [n]$. $$K_0(M) := F(M) / R(M)$$

For $\mathbb{N}$, this completion is precisely the integers, $\mathbb{Z}$.

Verse III: The Grothendieck Group of a Ring~
Let $R$ be a ring. We define $\mathbf{P}(R)$ to be the set of all isomorphism classes of finitely generated projective (left) modules over $R$.

Under the operation of direct sum ($\oplus$), $\mathbf{P}(R)$ forms an abelian monoid, where the isomorphism class of the zero module serves as the identity. We define the zeroth algebraic K-group of the ring $R$ as the group completion of this monoid: $$K_0(R) := \mathbf{P}(R)^{-1} \mathbf{P}(R)$$

Verse IV: The Eilenberg Swindle~
A crucial detail in the definition of $\mathbf{P}(R)$ is that the modules must be finitely generated. If we drop this finiteness condition, the theory collapses due to the Eilenberg Swindle.
Because $K_0(R)$ is a group, we can subtract $[R^\infty]$ from both sides, leaving us with $[P] = 0$. Consequently, every module becomes trivial, and $K_0(R) = 0$. The Eilenberg Swindle elegantly demonstrates that K-theory yields meaningful invariants exclusively for finite data.

Let $R^\infty$ denote a countably infinitely generated free module over $R$. Let $P$ be any finitely generated projective module. By definition, there exists some module $Q$ such that $P \oplus Q \cong R^n$ for some $n \in \mathbb{N}$.
When we take the direct sum of $P$ and $R^\infty$: $$\begin{aligned} P \oplus R^\infty &\cong P \oplus (Q \oplus P) \oplus (Q \oplus P) \oplus \cdots \\ &\cong (P \oplus Q) \oplus (P \oplus Q) \oplus \cdots \\ &\cong R^n \oplus R^n \oplus \cdots \\ &\cong R^\infty \end{aligned} $$
Passing this relation into the Grothendieck group yields: $$[P] + [R^\infty] = [R^\infty]$$

Verse V: Linear Algebra and Commutative PIDs~
Consider a field $k$. The finitely generated projective modules over $k$ are precisely the finite-dimensional vector spaces. When we take the direct sum of two vector spaces $V$ and $W$, their dimensions are additive: $$\dim(V \oplus W) = \dim V + \dim W$$
Well, this formalizes a familiar concept from linear algebra. For any linear map $T: V \to W$, the Rank-Nullity Theorem states: $$\dim V = \dim(\ker T) + \dim(\operatorname{im} T)$$ This demonstrates that dimension is the additive invariant.
For example, if $R$ is a Dedekind domain (ubiquitous in algebraic number theory), $K_0(R) \cong \mathbb{Z} \oplus \operatorname{Pic}(R)$, where $\operatorname{Pic}(R)$ is the ideal class group, which is an invariant measuring the extent to which unique factorization fails in $R$.
Because finite-dimensional vector spaces over a field are classified entirely up to isomorphism by their dimension, the Grothendieck group simply tracks this integer rank. Therefore: $$K_0(k) \cong \mathbb{Z}$$
This result extends gracefully to any commutative Principal Ideal Domain (PID). If $R$ is a PID, every finitely generated projective $R$-module is free. Thus, it can be identified up to isomorphism by its rank, $R^n$. The Grothendieck group again yields $K_0(R) \cong \mathbb{Z}$.
Note that $K_0(R)$ is not always isomorphic to $\mathbb{Z}$.

Epilogue~
"I am not talking to you now through the medium of custom, conventionalities, nor even of mortal flesh: it is my spirit that addresses your spirit; just as if both had passed through the grave, and we stood at God's feet, equal,—as we are!"
—Charlotte Brontë, Jane Eyre

Grothendieck’s formulation of $K_0$ established a universal framework for extracting additive invariants. However, this was merely the foundational rung of a much larger ladder. Mathematicians such as Hyman Bass, John Milnor, and Daniel Quillen generalized this machinery to define higher algebraic K-groups ($K_1, K_2, K_n$), establishing deep connections between ring theory, topology, and geometry.

This post is adapted from a presentation originally given for the course MTH666A: Category Theory, taught by Prof. Amit Kuber. 

Sketching a Colimit

Prologue~
Said the Duck to the Kangaroo, 
'Good gracious! how you hop!
Over the fields and the water too,
As if you never would stop!
My life is a bore in this nasty pond,
And I long to go out in the world beyond...'
— Edward Lear, The Duck and the Kangaroo (1871)

As a continuation to the previous post, let us look at pushouts. However, we take a different approach this time, we first examine the general architecture of a colimit, and then see how the pushout emerges as a specific, beautiful instance of it.

Verse I: The Roads~
A diagram of type $J$ in a category $\mathscr{C}$ is a covariant functor: $ D : J \to \mathscr{C}$, where category $J$ is called the index category and the functor $D$ is called a $J-$shaped diagram.
Think of a diagram as the categorical version of an indexed family in Set theory. While a family of sets is just a collection, a diagram includes the relationships (morphisms) between those objects. It is a network of roads (morphisms) connecting cities (objects).
Here on, we will look at the functorial aspect of the diagram $D$, so for the sake of better understanding, we refer to $D$ as $F$.

Verse II: At the Crossroads~
If the diagram is the network of roads, the colimit is where they all converge.
We define a co-cone of a diagram $F: J \to \mathscr{C}$ as an object $N \in Ob(\mathscr{C})$ together with a family of morphisms $\psi_X : F(X) \to N$ for all $X \in Ob(J)$ such that for every morphism $f: X \to Y$ in $J$, we have $$\psi_Y \circ F(f) = \psi_X$$.

Colimit of a diagram $F: J \to \mathscr{C}$ is a co-cone $(L, \phi)$ of $F$ such that for any other co-cone $(N, \psi)$ of $F$, there exists a unique morphism $u: L \to N$ such that for all $X \in Ob (J)$, we have $$u \circ \phi_X = \psi_X$$

Some examples of colimits are disjoint unions, direct sums, coproducts, pushouts, or direct limits. Note that pushouts are a form of colimits, and similarly, pullbacks are a form of limits.
In other words, $L$ is the closest or best meeting point, any other meeting point $N$ is just a shadow of $L$.

Verse III: Squaring the Circle~
Ideally, we want to glue two objects together.
Now, given a diagram consisting of two morphisms $f: Z \to X$ and $g: Z \to Y$ with a common domain. This $Z$ acts as the seam or the overlap between $X$ and $Y$. We define a pushout consisting of an object $P$ slong with two morphisms $p_1 : X \to P$ and $p_2 : Y \to P$ such that the square commutes.
$(P,p_1, p_2)$ is universal with respect to the diagram, that is to say that for any other such triple $(Q, q_1, q_2)$, the following diagram commutes such that there exists a unique $u: P \to Q$, also making the diagram commute. 

This is the categorical way of squaring the circle. We take the separate paths $X$ and $Y$ and force them to reconcile. We are effectively gluing $X$ and $Y$ together along $Z$, identifying the image of $Z$ in $X$ with its image in $Y$.
Pushouts are also called fibered coproducts, fibered sums, amalagamated sums, or cocartesian squares.  
Similar to before, a pushout, if exists, is unique up to a unique isomorphism. As with all universal properties, a pushout (if it exists) is unique up to a unique isomorphism.

Verse IV: Through the Looking Glass~
If we step through the looking glass, the arrows reverse, and colimits become limits.
A cone to $F$ is an object $N \in \mathscr{C}$ together with a family of morphisms $\psi_X : N \to F(X)$ indexed by the objects $X \in J$, such that for every morphism $f : X \to Y$ in $J$, we have $$F(f) \circ \psi_X = \psi_Y$$
                                        
A limit of the diagram $F: J \to \mathscr{C}$ is a cone $(L, \phi)$ to $F$ such that for every cone $(N, \psi)$ to $F$, there exists a unique morphism $u: N \to L$ such that for all $X \in Ob (J)$, we have $$\phi_X \circ u = \psi_X$$ 

$u$ is called the mediating morphism. Where the colimit was a destination, the limit is a source. Where the pushout glues, the pullback (the limit form of the square) finds the common ground.

Epilogue~
So away they went with a hop and a bound,
And they hopped the whole world three times round;
And who so happy,—O who,
As the Duck and the Kangaroo?
— Edward Lear, The Duck and the Kangaroo (1871)

Tracing a Limit

Prologue~
"Except for the point, the still point,

There would be no dance, and there is only the dance."
— T.S. Eliot, (Burnt Norton) Four Quartets
We will dicuss pullbacks in an intuitive manner, as we also define it mathematically. An intuitive understanding of limit will also be established by the end. However, the definitions of limits, colimits and pushouts will be dicussed in later posts.

Verse I: The Set-up~
Let us begin by considering two functions defined on sets, having a common codomain, $f: X \to Y$ and $g: Z \to Y$. We obtain fig.1, which is (let's say) an incomplete square. We wish to complete the square by adding another vertex to the diagram, such that the obtained square commutes. 

Since our maps are defined from sets, we can take the cartesian product of the sets $X$ and $Z$ and complete the diagram (see fig.2), where $\pi_1$ and $\pi_2$ are the projection maps on $X$ and $Z$, respectively. 

Here, the choice of the cartesian product can be said to be rather natural.
But, does the square commute? Well, not yet. The square commuting would imply that the composition along the defined maps (whichever is valid), along each element would give us the same result.
That is, $f \circ \pi_1 (_)= g \circ \pi_2 (_)$, where $(_)$ is any element of $X \times Z$. This may or may not hold for all elements of the cartesian product so we resctify that using a consistency condition check and define the cartesian product along (say) $Y$ such that for all elements belonging to the subset of the cartesian product, we would have the diagram to be commutative. Mathematically, we will define it as,

$$ X \times_Y Z= \{(x,z) \in X \times Z \mid f(x)= g(z)  \}$$

Verse II: When in Rings~
Consider another setting but of commutative unital rings, where we have two ring homomorphisms $\alpha: A \to B$ and $\beta: C \to B$. We again would obtain a similar diagram as before (see fig. 3) and we wish to complete the square. Similar to above, we can consider the product of the rings- $A \times C$ and using our consistency condition check, we define a restriction on it, or rather construct a subring of the product ring: $ A \times_B C= \{ (a,c)\in A \times C \mid \alpha(a)= \beta(c) \} $.

We must also define the two accompanying morphisms to complete the square, $\beta' : A\times_B C \to A,$   $ \alpha' : A \times_B C \to C $ given by $\beta' (a,c) = a$ and $ \alpha' (a,c)= c$ for all $(a,c) \in A \times_B C$. This would automatically give us $\alpha \circ \beta' = \beta \circ \alpha'$.

Similarly, we can contruct a lot of similar (natural) examples, given two morphisms, but what exactly are we doing here?

Verse III: The Anatomy~

If we take the category Sets (a category where objects are sets and morphisms are functions), then example 1 defines a pullback (the restrictive cartesian product)  and similarly, in the category of C-Rings (i.e., commutative unital rings), the second example again deals with a pullback.
More generally and explicitly stated, a pullback of a given pair of morphisms, say $\alpha$ and $\beta$ consists of an object $P$ and two accompanying morphisms $p_1: P \to X$ and $p_2: P \to Z$ for which the figure obtained would commute. when talking about pullbacks, we must then have a triple- $(P, p_1, p_2)$ which must be universal with respect to the given diagram. Universal property essentially states that- for a given problem, there is a most universal solution $X$, such that any other solution $Y$ is connected to $X$ by a unique structure-preserving map.
So, for any other such triple $(Q, q_1, q_2)$, where $q_1: Q \to X$, and $q_2: Q \to Z$, are morphisms with $f \circ q_1= g \circ q_2$, there must be a unique $u: Q \to P$ such that $p_1 \circ u =q_1$ and $p_2 \circ u=q_2$.  

Since we have already established universality of the pullback, it must be intuitive to realise that if a pullback exists, it is then unique up to a unique isomorphism. 
Pullbacks are a limit of the diagram, and are also refereed to as fiber products. Whereas, pushouts are a colimit of the diagram. We will look into limits, colimits and pushouts in upcoming expository discussions.

Epilogue~

“The only true voyage of discovery . . . would be not to visit strange lands, but to possess other eyes, to behold the universe through the eyes of another, of a hundred others, to behold the hundred universes that each of them beholds, that each of them is.”

— Marcel Proust, The Captive / The Fugitive

Talking of universality, Anveshanā has realeased its January issue of volume 2 (2026) which is available to read at this url.

A Category Theory (Mini)Tour

Prologue~
“To see a World in a Grain of Sand
And a Heaven in a Wild Flower,
Hold Infinity in the palm of your hand
And Eternity in an hour.”
— William Blake, Auguries of Innocence
With this post we delve into the very basics of a category and functors. We understand the adjunction between the $\mathrm{Hom}$ functor and the Tensor product- something which is quite interesting since it reveals to you more than you ask of it. There are several adjunctions which arise naturally, similar to this one and by the uniqueness property they can be shown to be isomorphic to each other, which is a very interesting result (if you ask me)!

Verse I: Categories~
A category $\mathcal C$ consists of the following:
1. A class of objects denoted as $ \mathrm{Ob}(\mathcal C)$;
2. for $X,Y \in  \mathrm{Ob}(\mathcal C)$, a class of morphisms from $X$ into $Y$ denoted as $\mathrm{Mor}_{\mathcal C}(X,Y)$;
3. for each $X, Y, Z \in  \mathrm{Ob}(\mathcal C),$ a composition map $$ \mathrm{Mor}_{\mathcal C} (X,Y) \times  \mathrm{ Mor}_{\mathcal C} (Y,Z) \to  \mathrm{Mor}_{\mathcal C} (X, Z)  $$ satisfying the associative property for all $ f \in  \mathrm{Mor}_{\mathcal C} (X,Y), g \in  \mathrm{Mor}_{\mathcal C} (Y,Z)$ and $h \in  \mathrm{Mor}_{\mathcal C}(Z,W)$, $(h \circ g) \circ f =  h \circ (g \circ f)$ holds, for every $X,Y,Z,W \in  \mathrm{Ob}(\mathcal C).$
4. Identity morphisms id_X ∈ Mor_C(X,X) for each X, satisfying f ∘ id_X = f = id_Y ∘ f.

Verse II: Monoids~

For fun's sake, let us recall the definition of a monoid- A monoid is a set $M$ with a binary operation, which is associative and has an identity. We can also state the same definition as- a monoid being a category with one object. Thus for any category $\mathcal C$ and any object $a \in  \mathrm{Ob}(\mathcal C)$, the class $ \mathrm{Mor}_{\mathcal C}(a,a)$ of all morphisms from $a$ to $a$ is a monoid. Essentially, the single object’s endomorphisms form a monoid under composition, with identity being the identity morphism.

Verse III: Functors~ A functor is a morphism between categories. Consider Categories $\mathcal A$ and $\mathcal B$, then a functor, say $F$ is defined as, $$F: \mathcal A \to \mathcal B$$ with domain $\mathcal A$ and the codomain $\mathcal B$, consisting of two related functions, namely the object function and the arrow function. The object function assigns each object  $a \in  \mathrm{Ob}(\mathcal A)$  with an object  $Fa$  of  $\mathcal B$  and the arrow function assigns to each arrow, say  $f: a \to a'$  of $\mathcal A$,  an arrow  $Ff: Fa \to Fa'$  of  $\mathcal B$.
$F$  hence carries each identity of  $\mathcal A$  to an identity of  $\mathcal B$  and each composable pair  $<g,f>$  in  $\mathcal A$  to a composable pair  $<Fg, Ff>$  in  $\mathcal B$  with  $Fg \circ Ff= F(g \circ f)$.

Verse IV: Hom bifunctor~
Now, consider the set of all morphisms from the definition of a category and let us define it functorially, i.e., we consider Hom to be a functor and define it as
$$ \mathrm{Hom}_{\mathcal C} ( - ,-) :=  \mathrm{Mor}_{\mathcal C} (-,-)$$ Let  $\mathcal C$  be a category. For objects  $A,B \in  \mathrm{Ob}(\mathcal C)$,  the set of morphisms from  $A$  to  $B$  is denoted by  $\mathrm{Mor}_{\mathcal C}(A,B)$.  Then the assignment  $(A,B) \longmapsto \mathrm{Mor}_{\mathcal C}(A,B)$  defines a bifunctor $$\mathrm{Hom}_{\mathcal C}(-,-) : \mathcal C^{\mathrm{op}} \times \mathcal C \to \mathbf{Set}$$ whose variance is described as follows:
1. Contravariance in the first argument.
Let  $f : A' \to A$  be a morphism in  $\mathcal C$,  and let  $B$  be a fixed object. Define a function $$\mathrm{Hom}_{\mathcal C}(A,B) \to \mathrm{Hom}_{\mathcal C}(A',B)$$ by  $h \longmapsto h \circ f .$  This assignment satisfies $$\mathrm{id}_A^\ast = \mathrm{id}_{\mathrm{Hom}(A,B)}, \qquad (g \circ f)^\ast = f^\ast \circ g^\ast, $$ and therefore defines a contravariant functor in the first variable.
2. Covariance in the second argument.
Let  $g : B \to B'$  be a morphism in  $\mathcal C$,  and let  $A$  be a fixed object. Define a function $$\mathrm{Hom}_{\mathcal C}(A,B) \to \mathrm{Hom}_{\mathcal C}(A,B')$$ by  $ h \longmapsto g \circ h .$  This assignment satisfies $$\mathrm{id}_B{}_\ast = \mathrm{id}_{\mathrm{Hom}(A,B)}, \qquad (g' \circ g)_\ast = g'_\ast \circ g_\ast,$$ and therefore defines a covariant functor in the second variable.
This is exactly what the  $\mathrm{Hom}$  functor does, and hence it is contravariant in its first argument and covariant in its second, and consequently defines a bifunctor $$\mathrm{Hom}_{\mathcal C}(-,-) : \mathcal C^{\mathrm{op}} \times \mathcal C \to \mathbf{Set}.$$ Note that in the first argument, $ \mathrm{Hom}$ admits an object from the opposite category ${\mathcal C}^{op}$.  

Verse V: Tensor Product~
Let  $R$  be a commutative ring, and let  $\mathbf{Mod}_R$  denote the category of  $R$-modules.
For  $R$-modules  $M$  and  $N$ , the tensor product  $M \otimes_R N$  is an  $R$-module equipped with a bilinear map $$\tau : M \times N \to M \otimes_R N$$ satisfying the following universal property.
For every  $R$-module  $P$, composition with  $\tau$  induces a bijection $$\mathrm{Hom}_R(M \otimes_R N, P) \;\cong\; \mathrm{Bil}_R(M \times N, P),$$ where  $\mathrm{Bil}_R(M \times N, P)$  denotes the set of  $R$-bilinear maps from  $M \times N$  to  $P$.
Equivalently, for every bilinear map  $\beta : M \times N \to P$  there exists a unique  $R$-linear map $\widetilde{\beta} : M \otimes_R N \to P,$   such that  $\beta = \widetilde{\beta} \circ \tau .$
The tensor product defines a bifunctor $$-\otimes_R- : \mathbf{Mod}_R \times \mathbf{Mod}_R \to \mathbf{Mod}_R $$ which is right exact in each variable.
Fixing an  $R$-module $N$, the assignment  $M \longmapsto M \otimes_R N$  defines a covariant functor $$-\otimes_R N : \mathbf{Mod}_R \to \mathbf{Mod}_R.$$ Tensors products and Hom functors are bifunctors in a monoidal category.

Verse VI: Adjunctions~

Let $\mathcal C$ and $\mathcal D$ be categories, and let  $F : \mathcal C \to \mathcal D$  and  $G : \mathcal D \to \mathcal C $  be functors.

The functor  $F$  is said to be left adjoint to  $G$ , and  $G$ is said to be right adjoint to  $F$ , if for every object $C \in \mathcal C$  and every object  $D \in \mathcal D$  there exists a bijection $$\mathrm{Hom}_{\mathcal D}(F(C), D)  \;\cong\; \mathrm{Hom}_{\mathcal C}(C, G(D)).$$

This bijection is required to be natural in both variables, that is, natural in  $C$  and natural in  $D$. Equivalently, the adjunction is specified by a pair of natural transformations
$$\eta : \mathrm{id}_{\mathcal C} \to G \circ F \quad \text{and} \quad \varepsilon : F \circ G \to \mathrm{id}_{\mathcal D},$$ called the unit and counit of the adjunction, satisfying the identities $$(\varepsilon \circ F) \circ (F \circ \eta) = \mathrm{id}_F, \qquad (G \circ \varepsilon) \circ (\eta \circ G) = \mathrm{id}_G.$$ For an adjunction  $F \dashv G$  between categories  $\mathcal{C}$ and $\mathcal{D}$ , the unit and counit are (said to be) natural transformations $$\eta : \mathrm{id}_{\mathcal{C}} \longrightarrow G \circ F, \qquad \varepsilon : F \circ G \longrightarrow \mathrm{id}_{\mathcal{D}}.$$ Explicitly, for each object  $C \in \mathcal{C}$  and  $D \in \mathcal{D}$  there are morphisms $$ \eta_C : C \longrightarrow G(F(C)), \qquad \varepsilon_D : F(G(D)) \longrightarrow D, $$ which are natural in  $C$  and  $D$ , respectively. These satisfy the triangle identities. For every object  $C \in \mathcal{C}$ , $$ F(C) \xrightarrow{F(\eta_C)} F(G(F(C))) \xrightarrow{\varepsilon_{F(C)}} F(C)$$ is equal to  $\mathrm{id}_{F(C)}$;  and for every object  $D \in \mathcal{D}$,  $$ G(D) \xrightarrow{\eta_{G(D)}} G(F(G(D))) \xrightarrow{G(\varepsilon_D)} G(D)$$ is equal to  $\mathrm{id}_{G(D)}$.

$(\varepsilon_{F(-)} \circ F(\eta_-)) = \mathrm{id}_F$ and $(G(\varepsilon_-) \circ \eta_{G(-)}) = \mathrm{id}_G$

These identities express the coherence of the adjunction: going from  $F(C)$  to  $G(F(C))$  via the unit and back via the counit does nothing overall, and similarly for  $G(D)$.

Verse VII: Hom-⊗ Adjunction~

Let $R$ be a commutative ring. For $R$-modules $M,N,P$, there is a natural isomorphism $$\mathrm{Hom}_R(M \otimes_R N, P) \;\cong\; \mathrm{Hom}_R\bigl(M,\mathrm{Hom}_R(N,P)\bigr). $$ One should also note the symmetric role of $M$ and $N$, that is we finally obtain the following, $$\mathrm{Hom}_R(M \otimes_R N, P) \;\cong\; \mathrm{Hom}_R\bigl(M,\mathrm{Hom}_R(N,P)\bigr) \;\cong\; \mathrm{Hom}_R(N \otimes_R M, P)$$ For fixed $N, \ − ⊗_R N$ is left adjoint to $\mathrm{Hom}_R(N, −)$, with  $Φ$  as the adjunction isomorphism.

Epilogue~
“For last year's words belong to last year's language
And next year's words await another voice.”
— T. S. Eliot

One can refer to Categories for the Working Mathematician by Saunders Maclane, for a detailed study. 

This post is adapted from a presentation originally given for the course MTH619: Representation Theory of Quivers, taught by Prof. Amit Kuber. 

Of Algebra and Diagrams...

Prologue~
“Algebra is generous; she often gives more than is asked of her.”
— Jean le Rond d'Alembert

The aim of this post is to discuss a few theorems based of ring homomorphisms, while giving an intuition of diagram chasing, since our magic tool here would be commutative diagrams. We also encounter what a natural map would be in such a setting.

Verse I: The Epimorphism Theorem~

Let $f : R \to S$ be a ring homomorphism then there exists an isomorphism from $R/ Ker\ f$ to $S$, iff $f$ is an epimorphism.

Fig. 1

Now if you consider the adjacent figure, then clearly $f$ has been rightfully depicted as a map from $R$ to $S$.

We were supposed to find the map between the quotient ring and $S$, which here is represented as $\tilde{f}$.

Notice that the map denoted by $\eta$ here is a natural (or canonical) surjection, and it is always a surjective ring homomorphism. 

The meaning of surjective here is simply onto, i.e., the map $\eta$ takes an element of $R$ and lands it in $R/Ker \ f$, such that every element of $R/Ker\ f$ has a pre-image in $R$.

Now we must define $\tilde{f}$ to complete the proof. So what our $\tilde{f}$ does here is takes an element from $R/Ker \ f$ and maps it to the image of $R$ under $f.$

Hence, the above paragraph(s), for all elements $x \in R$ can be concluded as, 

$f(x)= \tilde{f} \circ \eta (x)$ 

and we say that the diagram in Fig. 1 commutes!

Here $\circ$ implies pre-right composition.

Verse II: Quotient of a Quotient Theorem~

If $I \subseteq J$ are both 2-sided ideals in $R$, then $(R/I)/(J/I)$ is naturally isomorphic to $R/J.$

Fig. 2

Let us dissect the adjacent figure which, one can claim that gives a precise and swift proof of the theorem stated above.

Here again, we have two quotient rings $R/I$ and $R/J$, where our $I$ and $J$ are ideals. And we have been given that $I \subseteq J$. (In)formally, our $I$ sits in $J$ and hence from what we did in the previous proof, we can say that the map $\eta_{IJ}$ forms a natural (or canonical) surjection, which again means that for every element of $R/I$, there exists a pre-image in $R$. 

Similar applies for $\eta_J$ and the map here is (again) a canonical surjection.

Interestingly, $\eta_{R/J}$ and $\eta_{IJ}$ are also canonical surjections from their respective domains to their codomains.

Fig. 2.1

Now, talking about the map which actually concerns us, because everything so far arose naturally from the ideal and quotient ring relations. 

We are interested in $\tilde{\eta}$, which is a map from $(R/I)/(J/I)$ to $R/J$, and since $I \subseteq J$, this again turns out to be another natural surjection, the only difference being that it's slightly more demanding than the others.

What this map does is take an element of the form $x+I+J/I$, where $x \in R$ and maps it to an element in $R/J$ of the form $x+J$, which is obvious, since we started with $x+I+J/I= x+I + j+I , \  \forall j \in J$

and since $I \subseteq J$, $j+I, \ \forall j \in J,$ belongs to $J$. Therefore, we can concisely write the $  J+I = J$.

Hence, $x+I+J/I = x+I + j+I= x+I+J= x+J $.

And our Fig. 2 commutes, which implies that $  \eta_I \circ \eta_{IJ}= \eta_J $ and $\eta_{IJ}= \tilde{\eta} \circ \eta_{R/ J}$

One may verify these.

Also see that we can combine the other two relations and write 

$ \eta_I \circ (\tilde{\eta} \circ \eta_{R/ J} )= \eta_J $

Verse III: Extension and Ideal Theorem~

If $I$ is a 2-sided ideal in $R$, then $I[X]$ is a 2-sided ideal of $R[X]$ and the quotient ring $R[X]/I[X]$ is naturally isomorphic to $(R/I)[X]$.

Fig. 3

The first part of the theorem can be proved by considering $I$ to be the 2-sided ideal and then we can proceed by writing $I[X]$ as $$ I[X]=\{  a_0 +a_1 X + a_2 X^2 + \cdots + a_r X^r , \ \forall \ a_i \in I \} $$

And, $I[X]$ is a 2-sided ideal in $R[X]$.

Now, consider the adjacent figure. 

Here, the maps $\gamma_{R/I}$ and $\gamma_{I[X]}$ are both natural surjections.

And our $\gamma_{R/I}$ is defined by 

$a_0 +a_1 X + a_2 X^2 + \cdots + a_r X^r  = \overline{a}_0 + \overline{a}_1 X + \overline{a}_2 X^2 + \cdots + \overline{a}_r X^r  $

where $\overline{a}= a+I, \ \forall \  a \in R$.

So, we are now concerned with our map $\tilde{\gamma}$ which takes objects from $R[X]/I[X] $ to an object in $(R/I)[X]$. Clearly, the objects of the domain are of the form 

$\tilde{\gamma}(a(X) + I[X]) = (a_0 + I) + (a_1 + I)X + \cdots + (a_n + I)X^n$

 and clearly, $(a_0 + I) + (a_1 + I)X + \cdots + (a_n + I)X^n \ \in (R/I)[X]$ and it's a surjection. every element of $(R/I)[X]$ has a pre-image in $R[X]/I[X] $.

Verse IV: Quotient maps~

Fig. 4

In all the above commutative diagrams, we obtain a figure of the adjacent form, and if all the maps in the figure, i.e., $\alpha, \beta$ and $\delta$ are onto, then the map $\alpha$ is said to factor through the object $C$ such that $ \alpha = \delta \circ \beta$ and hence in all the above figures, we obtain this universal property.

In general, a universal property describes how a particular object (such as a quotient ring, product, coproduct, or free object) is characterized by its relationships to other objects via maps that uniquely factor through it. 

Note that the diagram in Fig. 4 commutes.

Epilogue~

Algebra's real generosity is in its ability to unify results.

Many fundamental theorems, such as the isomorphism theorems for ring homomorphisms, become clear and accessible when approached through the lens of commutative diagrams and natural maps. These tools reveal underlying connections and provide a systematic framework for understanding why these results hold and how they fit together.

Reference- C. Musili, Rings and Modules.

On Siken and Being~

Richard Siken’s writing has become a touchstone, especially the book I recently devoured- The War of the Foxes. Because of the way he blurs certainty and leaves the reader suspended between clarity and disarray. His poems often weave images of intimacy with undercurrents of danger, collapsing the distinct between love and violence, desire and loss.
Inspired by him, I tried to write something, titled- Being and the Blur.

I took a deep breath and let it out.

What was exhaled was no more a breath,
instead an air that mixes with the atmosphere.
We call it life.

Something you give out is life.
Something you have to hold in is death.
And if you hold everything in,
maybe you are dead.

I too took my anger and created a demon out of it.
I call that demon an angel.
You look too long and the lines begin to blur.
What is a 'land' and what is a 'sea'?
Do you see?

The see glass is mint green.
Green- the color of jealousy they say.
I see trees of green and not an ounce of hatred.

We won't lead anywhere in a world like this.

~Purnima

In search of a nicer $G$...

Prologue~
“What is it indeed that gives us the feeling of elegance in a solution, in a demonstration? It is the harmony of the diverse parts, their symmetry, their happy balance; in a word it is all that introduces order, all that gives unity, that permits us to see clearly and to comprehend at once both the ensemble and the details.”
—From Science et Méthode (1908), Livre Premier, Chapitre 2, p. 25; Translated in The Foundations of Science: Science and Hypothesis, The Value of Science, Science and Method (1913), p. 372. by  Henri Poincaré

Verse I: What is a 'nicer' G?
So I have a group $G$- (say, a non-trivial one) which can either have a commutative or non-commutative structure, that is, be abelian or non-abelian. Now, abelian groups are very 'nice' to deal with- they make things easier for us, but non-abelian groups, not so much...
So how can one make things easier in a non-abelian group, or rather- how can we work in or deal with a 'nice' part of this non-abelian group which can make things (again, say-) 'nicer'? We explore this question in this post.

Verse II: The canvas~
Let $G$ be any non-trivial group, then if $a,b \in G$, then the commutator of $a$ and $b$ is the element $ab a^{-1} b^{-1}$.  
A good observation would be to see that if our $G$ is abelian, then our commutator is simply the identity element $e \in G$. Now, let us define $C$ to be the set\[ C =\{ x_1 x_2 x_3 \cdots x_n \ | \ n \geq 1, \text{each $x_i$ is a commutator in $G$} \} \] Basically, our $C$ here is the collection of all finite products of commutators in $G$. One can intuitively see that $C$ then will be a normal subgroup of $G$. \[ C \lhd G\] If not, maybe we can try thinking about it- as mentioned above, at least the identity element is a commutator and hence our $C$ is non-empty. And given any two elements in $C$, they are of the form $x_1 x_2 \cdots x_n$ and $y_1 y_2 \cdots y_n$ and since this is just a finite product of commutators, we also have their inverses existing, which then again belong to $C$. Now we can take an element $g \in G$ and apply the left conjugation action by $g$ on an element of $C$. As depicted below, we can play around!\[ g c g^{-1} = g x_1 x_2 \cdots x_n g^{-1} = (g x_1 g^{-1}) (g x_2 g^{-1}) \cdots g(x_n g^{-1}) \] Also notice that given any arbitrary element $c \in C$, we have, \[ (g c g^{-1})^{-1} = g c^{-1} g^{-1}\] and we obtain our equation back. Moreover, $g c g^{-1}$ was a product of commutators and hence it belongs in $C$.
$\therefore$ $\forall \ g \in G$ and $c \in C$, $gcg^{-1} \in C$, and hence, $C \lhd G$.
The normal subgroup $C$ of $G$ is  called the commutator subgroup or the derived subgroup of $G$ and is usually denoted by $C= G' = [G,G]$.
Also, if our $G$ is abelian (to begin with,) we then obtain $C=\{e\}$ and hence one can see the commutator subgroup as one measure of how far away a group is from being abelian.
To answer the question we began with, what does it mean to abelianize a group?

Verse III: Towards Abelianization~
What it intuitively means is that whenever $g,h \in G$, we must have $hg=gh$ with respect to the group operation (which has been suppressed here). But from the sketch of the proof done above, we realise that an abelian group has $C=\{e\}$ and therefore, in particular, we want every element of the form $ghg^{-1} h^{-1}$ to be the identity. And indeed, this is all that we need to do!
That is to say, if $gh g^{-1} h^{-1} = e, \ \forall g,h \in G$, then our $G$ is abelian!
We can see this as, \[ ghg^{-1}h^{-1}= e \Rightarrow ghg^{-1}= h  \Rightarrow gh =hg \]
What does this tell us though?
If we make every commutator of our group $G$ trivial, then we will abelianize $G$, and conversely, to make $G$ abelian, we have to make every commutator of our group $G$ trivial.
But, how do we do that?
We have a commutator subgroup $C$ which is normal in $G$. We can take the quotient of $G$ by $C$ (since it will give us a quotient group) and it makes good sense as the first step to play around!
Now, what we obtain will be a quotient group consisting of all the cosets of $C$ in $G$ and one can see that this very well could be (rather will be) the 'nicer', that is, the 'abelian version' of our group $G$ which we have been looking for. Now let's see if this works!
Consider the abelianization of $G$: \(G^{\mathrm{ab}} = G/[G,G]\)
Consider the quotient \[G^{\mathrm{ab}} \;:=\; G/[G,G]\] It is abelian because for any \(a,b \in G\), which is to say that, \[(a[G,G])\,(b[G,G]) \;=\; ab[G,G]\] while \[(b[G,G])\,(a[G,G]) \;=\; ba[G,G]\] But \(ab\) and \(ba\) differ by the commutator- what does that mean?
Well, since $ab= [a,b] (ba)$, the two cosets coincide. 
We have $[a,b] = aba^{-1} b^{-1}$ and if we multiply on the right  by $(ba)$, we obtain \[ [a,b] (ba)= aba^{-1}b^{-1} (ba) = ab a^{-1} (b^{-1} b) a = ab a^{-1} a = ab\] \[\therefore ab = [a,b](ba),\] so \[ab[G,G] = [G,G] (ba)\] since \([a,b]\in [G,G]\). Hence the cosets commute and \(G^{\mathrm{ab}}\) is abelian. Conversely, if every commutator is trivial in a quotient \(G/N\), then \([G,G]\subseteq N\). This gives the minimality of \([G,G]\): \([G,G]\) is the smallest normal subgroup of \(G\) with abelian quotient, i.e. for any normal \(N\trianglelefteq G\), \(G/N\) is abelian if and only if \([G,G]\subseteq N\).

Verse IV: Universality in Action~
Now, the abelianization of our group $G$ is not just an abelian quotient; it is the universal one.
If \(A\) is abelian and \(\varphi:G\to A\) is any homomorphism, then there exists a unique homomorphism \(\overline{\varphi}:G^{\mathrm{ab}}\to A\) such that \(\varphi = \overline{\varphi}\circ \pi\), where \(\pi:G\to G^{\mathrm{ab}}\) is the natural projection.
Why? Since \(A\) is abelian, \(\varphi([a,b])=e\) for all \(a,b\in G\), so \([G,G]\subseteq \ker\varphi\).
And by the First Isomorphism Theorem, \(\varphi\) factors uniquely through \(G/[G,G]\).
And guess what? This universal property is often the most efficient definition of abelianization.

Epilogue~
When we begin with a non-abelian group, we constantly worry about the turbulence of non-commutativity and what abelianization does for us is- it takes away that turbulence and offers us a 'nicer' environment to work with. 
The process gives a precise answer: we take the quotient by the commutator subgroup, leaving the 'largest' abelian image of the original group. This is not just convenient—it is universal. Any map from our group to an abelian group must factor uniquely through this construction. That is why abelianization matters: it formalizes the idea of extracting commutativity in the most general and economical way, without discarding more structure than necessary.

References- (1), (2).