[Math] Proving that pullback objects are unique up to isomorphism

abstract-algebracategory-theory

In Hungerford's Algebra he defines a pullback of morphisms $f_1 \in \hom(X_1,A)$ and $f_2 \in \hom(X_2,A)$ as a commutative diagram
$$\require{AMScd}
\begin{CD}
P @>{g_1}>>X_2\\
@V{g_2}VV @V{f_1}VV \\
X_2 @>{f_2}>> A
\end{CD}$$
satisfying the universal property that for any commutative diagram
$$\require{AMScd}
\begin{CD}
Q @>{h_1}>>X_2\\
@V{h_2}VV @V{f_1}VV \\
X_2 @>{f_2}>> A
\end{CD}$$
there exists a unique morphism $t: Q \to P$ such that $h_i = g_i \circ t$. He then asks the reader to establish that

For any other pullback diagram with $P'$ in the upper-left corner $P \cong P'$.

How do we obtain this isomorphism?

The obvious choice seems to be considering the two morphisms $t: P \to P'$, $t': P' \to P$ and show that they compose to the identity. To this end,

$$h_1 = g_1 \circ t \implies h_1\circ 1 = h_1 \circ t' \circ t$$

but we cannot cancel unless $h_1$ is monic. Can we claim that necessarily $t \circ t'$ is the identity, since comparing $(P,g_1,g_2)$ with itself there exists a unique morphism $t'': P \to P$?

Best Answer

As said by Martin Brandenburg in the comments, you stated the universal property wrong (not relevant anymore since the edit of the original post).

A pullback of $f_1\colon X_1 \to A,f_2\colon X_2 \to A$ is a diagram $$ \require{AMScd} \begin{CD} P @>{g_1}>>X_1\\ @V{g_2}VV @V{f_1}VV \\ X_2 @>{f_2}>> A \end{CD} $$ satisfying that for any other diagram $$\require{AMScd} \begin{CD} Q @>{h_1}>>X_1\\ @V{h_2}VV @V{f_1}VV \\ X_2 @>{f_2}>> A \end{CD}$$ there exists a unique $t \colon Q \to P$ such that the following diagram commutes :

                                                           enter image description here .

So now, if you have two pullback $P,P'$ there is $t \colon P \to P',t' \colon P' \to P$ such that commute the following diagrams :

                               enter image description here                            enter image description here .

Notably, the arrow $t' \circ t \colon P \to P$ make the diagram

                                                           enter image description here

commutes. By the universal property of the pullback $P$, such a $t'\circ t$ is unique : do you see another arrow $P \to P$ satisfying the same property ? Then it must equal $t' \circ t$.

Starting from here and elaborating a similar argument with the pullback $P'$, you should be able to prove the uniqueness up to isomorphism.

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