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Let $X$ be $Top(X)$ be the category of open sets of $X$ with inclusion maps as morphism. Let $\mathcal{C}$ be abelian category and $\mathcal{C}_x$ denote the category of contravariant functors from $Top(X)$ to $\mathcal{C}$. Let $\mathcal{F}$ and $\mathcal{G} \in obj(\mathcal{C}_x)$. I want to show that product of $\mathcal{F}$ and $\mathcal{G}$ exist. Let us $\mathcal{F} \times \mathcal{G}(U)=\mathcal{F}(U) \times \mathcal{G}(U)$. I want to show this is the direct product. Let $\mathcal H \in \mathcal{C}_x$ with natural tranformations $i_1$ and $i_2$ to $\mathcal{F}$ and $\mathcal{G}$. Now, I know there is unique natural transformation $\eta$ from $\mathcal H$ to $\mathcal{F} \times \mathcal{G}$ as there is a unique map from $\mathcal H(U)$ to $\mathcal{F}(U) \times \mathcal{G}(U)$ for any open set $U$. But how do I show that naturality square for $\eta$ commutes?

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Commutes with the restriction maps? –  Loki Clock Jun 21 '13 at 17:01
    
Yes with the restriction maps –  Mohan Jun 21 '13 at 17:02

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By the naturality square you mean the diagram you get with respect to some inclusion $V \to U$? If so then start with $x \in \mathcal H(U)$, push $x$, using $i_1$ and $i_2$ to $y \in \mathcal F(U)$ and $z \in \mathcal G(U)$. By naturality of $i_1$ and $i_2$ pushing $x|_V$ through $i_1$ and $i_2$ yields $y|_V$ and $z|_V$.

Now by definition of the maps $\mathcal H(U) \to \mathcal F(U) \times \mathcal G(U)$ we have that $x$ goes to $(y, z)$ and $x|_V$ goes to $(y|_V, z|_V)$ and by definition of the restrictions of $\mathcal{F \times G}$ we get that $(y, z)|_V = (y|_V, z|_V)$, thus $\mathcal{H \to F \times G}$ is natural.

You should definitely draw out a big diagram on paper and trace through what I've said here to make sure you understand it because it can be confusing at first, but once you get it you'll see that it amounts to nothing more than a diagram chase.

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Yes, but I was trying to prove it in a general abelian category. –  Mohan Jun 21 '13 at 17:08
    
Well, you can either use the Freyd-Mitchell embedding theorem to say that a diagram chase works in a general category or you can use MacLanes diagram chase rules to say it does, or you can do it using the universal property of a product. –  Jim Jun 22 '13 at 5:11
    
general abelian category, that is. –  Jim Jun 22 '13 at 19:13

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