Prologue~
"Except for the point, the still point,
"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.
— 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,
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) \} $.
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~
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~
— 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.
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 $p_2 \circ u=q_2$.
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.”
Talking of universality, Anveshanā has realeased its January issue of volume 2 (2026) which is available to read at this url.