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 $t'': P \to P$?

  • $\begingroup$ 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). $\endgroup$ – Andreas Blass Aug 7 '13 at 4:32
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    $\begingroup$ The universal property is not correct as stated. There is a unique morphism $t : Q \to P$ making the diagrams commute(!). $\endgroup$ – Martin Brandenburg Aug 7 '13 at 7:37

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).


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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