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Given a category $C$, $X$,$Y$,$Z \in ObC$, $v\in Hom_C (X,Z)$, and $h \in Hom_C (Y,Z)$, let $(P,h',v')$ be their pullback. Given $H \in C$ and two morphisms $f,g \in Hom_C (H,P)$, $(H,v'\circ f,h'\circ g)$ is another such pullback. By the universal property of pullbacks, there is a unique morphism, $u\in Hom_C(H,P)$ such that the relevant diagram commutes. However because $f,g$ are two such morphisms it seems to me that this implies $f=g$. This seems wrong to me because $P$ and $H$ are just any objects and there is no reason I can see why they can't have more than one morphism between them. Unfortunately I can't see the flaw in my argument. Sorry I didn't draw a commutative diagram with this question, I will try to add it later.

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    $\begingroup$ You need to assume $vv'f=hh'g$, in which case it does follow that $f=g$. $\endgroup$ – Kevin Carlson Jan 21 '17 at 1:26

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