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I am trying to prove that the hom-functor $Hom(A,-)$ preserves pullbacks. I stuck at showing uniqueness. Could you please give me some hints?

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It preserves all limits. Try showing that instead. –  Zhen Lin Dec 8 '13 at 16:07
@ZhenLin, well, I still do not know limits (it is the next chapter in our lecture notes), so it supposed that I can show this without knowledge about limits :) –  user112072 Dec 8 '13 at 16:09

1 Answer 1

Consider some pullback $X \times_S Y$ in a category. In order to show that $\hom(A,-)$ preserves it, we have to show that the natural map $$\hom(A,X \times_S Y) \to \hom(A,X) \times_{\hom(A,S)} \hom(A,Y)$$ is an isomorphism. By construction of pullbacks in $\mathsf{Set}$, the right hand side is simply the set of pairs of maps $A \to X$ and $A \to Y$ such that the diagram $$\begin{array}{c} A & \rightarrow & X \\ \downarrow && \downarrow \\ Y & \rightarrow & S \end{array}$$ is commutative. Hence, the map is bijective by definition of a pullback!

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(In particular don't waste your time in proving existence+uniqueness, show bijectivity directly) –  Martin Brandenburg Dec 8 '13 at 16:15
Sorry, it turns out that I have never seen the notation $X \times_S Y$ for pullbacks. Is it just something like $X \rightarrow S \leftarrow Y$? –  user112072 Dec 8 '13 at 16:19
$X \times_S Y$ is the pullback of $X \to S \leftarrow Y$ (see any book, Wikipedia article, etc.) How do you denote the pullback? –  Martin Brandenburg Dec 8 '13 at 16:22
And $hom(A,X \times_S Y)$ is just application of this functor to all elements in a pullback? –  user112072 Dec 8 '13 at 16:25
Erm. Perhaps you should learn basic set theory and the definition of the hom functor before dealing with pullbacks. –  Martin Brandenburg Dec 8 '13 at 16:36

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