Andrej Bauer has a paper titled The pullback lemma in gory detail that goes over the proof of the pullback lemma in full detail. This is a basic result of category theory and most introductions leave it as an exercise. It is a good exercise, and you should prove it yourself before reading this article or Andrej Bauer’s.

Andrej Bauer’s proof is what most introductions are expecting you to produce. I very much like the representability perspective on category theory and like to see what proofs look like using this perspective.

So this is a proof of the pullback lemma from the perspective of representability.

The key thing we need here is a characterization of pullbacks in terms of representability. To
just jump to the end, we have for |f : A \to C| and |g : B \to C|, |A \times_{f,g} B| is **the
pullback of |f| and |g|** if and only if it represents the functor
\[\{(h, k) \in \mathrm{Hom}({-}, A) \times \mathrm{Hom}({-}, B) \mid f \circ h = g \circ k \}\]

That is to say we have the natural isomorphism \[ \mathrm{Hom}({-}, A \times_{f,g} B) \cong \{(h, k) \in \mathrm{Hom}({-}, A) \times \mathrm{Hom}({-}, B) \mid f \circ h = g \circ k \} \]

We’ll write the left to right direction of the isomorphism as |\langle u,v\rangle : U \to A \times_{f,g} B|
where |u : U \to A| and |v : U \to B| and they satisfy |f \circ u = g \circ v|. Applying
the isomorphism right to left on the identity arrow gives us two arrows |p_1 : A \times_{f,g} B \to A|
and |p_2 : A \times_{f,g} B \to B| satisfying |p_1 \circ \langle u, v\rangle = u| and
|p_2 \circ \langle u,v \rangle = v|. (Exercise: Show that this follows from being a *natural* isomorphism.)

One nice thing about representability is that it reduces categorical reasoning to set-theoretic reasoning that you are probably already used to, as we’ll see. You can connect this definition to a typical universal property based definition used in Andrej Bauer’s article. Here we’re taking it as the definition of the pullback.

The claim to be proven is if the right square in the below diagram is a pullback square, then the left square is a pullback square if and only if the whole rectangle is a pullback square. \[ \xymatrix { A \ar[d]_{q_1} \ar[r]^{q_2} & B \ar[d]_{p_1} \ar[r]^{p_2} & C \ar[d]^{h} \\ X \ar[r]_{f} & Y \ar[r]_{g} & Z }\]

Rewriting the diagram as equations, we have:

**Theorem**: If |f \circ q_1 = p_1 \circ q_2|, |g \circ p_1 = h \circ p_2|, and |(B, p_1, p_2)| is a
pullback of |g| and |h|, then |(A, q_1, q_2)| is a pullback of |f| and |p_1| if and only if
|(A, q_1, p_2 \circ q_2)| is a pullback of |g \circ f| and |h|.

**Proof**: If |(A, q_1, q_2)| was a pullback of |f| and |p_1| then we’d have the following.

\[\begin{align} \mathrm{Hom}({-}, A) & \cong \{(u_1, u_2) \in \mathrm{Hom}({-}, X)\times\mathrm{Hom}({-}, B) \mid f \circ u_1 = p_1 \circ u_2 \} \\ & \cong \{(u_1, (v_1, v_2)) \in \mathrm{Hom}({-}, X)\times\mathrm{Hom}({-}, Y)\times\mathrm{Hom}({-}, C) \mid f \circ u_1 = p_1 \circ \langle v_1, v_2\rangle \land g \circ v_1 = h \circ v_2 \} \\ & = \{(u_1, (v_1, v_2)) \in \mathrm{Hom}({-}, X)\times\mathrm{Hom}({-}, Y)\times\mathrm{Hom}({-}, C) \mid f \circ u_1 = v_1 \land g \circ v_1 = h \circ v_2 \} \\ & = \{(u_1, v_2) \in \mathrm{Hom}({-}, X)\times\mathrm{Hom}({-}, C) \mid g \circ f \circ u_1 = h \circ v_2 \} \end{align}\]

The second isomorphism is |B| being a pullback and |u_2| is an arrow into |B| so it’s necessarily of the form |\langle v_1, v_2\rangle|. The first equality is just |p_1 \circ \langle v_1, v_2\rangle = v_1| mentioned earlier. The second equality merely eliminates the use of |v_1| using the equation |f \circ u_1 = v_1|.

This overall natural isomorphism, however, is exactly what it means for |A| to be a pullback of |g \circ f| and |h|. We verify the projections are what we expect by pushing |id_A| through the isomorphism. By assumption, |u_1| and |u_2| will be |q_1| and |q_2| respectively in the first isomorphism. We see that |v_2 = p_2 \circ \langle v_1, v_2\rangle = p_2 \circ q_2|.

We simply run the isomorphism backwards to get the other direction of the if and only if. |\square|

The simplicity and compactness of this proof demonstrates why I like representability.