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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 $t'': P \to P$?

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As Zev wrote, the answer is yes, the uniqueness of $t''$ gives that $t'\circ t=1_P$. Don't forget that you also need $t\circ t'=1_{P'}$ (by the same method, of course). –  Andreas Blass Aug 7 '13 at 4:32
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The universal property is not correct as stated. There is a unique morphism $t : Q \to P$ making the diagrams commute(!). –  Martin Brandenburg Aug 7 '13 at 7:37
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2 Answers

up vote 2 down vote accepted

To answer the question at the very end of your post: yes. This sort of argument is fundamental, and applies in essentially the same way to any universal property (Wikipedia).

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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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