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In Mac Lane's, Categories for the Working Mathematician, on p.15 ex.3c) it asks to interpret "functor" when F: (group G)-->Set is a permutation representation of G.

Here is where I get stuck, G is one object category, so G gets mapped to one specific set, lets call it A, in the category Set. And each arrow of G gets mapped to a function from A to A? These functions must be bijections, in fact permutations, but how do we show that?

The definition of a permutation representation of a group G is h(g)(a)=g*a for g in G, a in A, but I do not know how to make sense of this definition in this context. Thanks.

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What you've written isn't a definition of a permutation representation. –  Qiaochu Yuan Sep 2 '11 at 1:24

1 Answer 1

up vote 5 down vote accepted

When you consider a group $G$ as a category $\mathcal G$, the category has one object $\star$ and the set of homomorphisms $\hom_{\mathcal G}(\star,\star)$ is $G$, with composition the multiplication in $G$. You can easily check that the identity morphism of $\star$ is the identity element in $G$, and that every morphism $\star\to\star$ in $\mathcal G$ is an isomorphism.

Now consider a functor $F:\mathcal G\to C$ where $C$ is any category. Then $F(\star)$ is some object $X$ in $C$, and for each arrow $g:\star\to\star$, we have a map $F(f):X\to X$. Now, since $F$ is a functor, the image under $F$ of every isomorphism in $\mathcal G$ is an isomorphism in $C$. It follows then that $F(f)$ is an isomorphism.

In the special case in which $C$ is the category $\mathbf{Set}$ of sets, an isomorphism is a bijection, so your statement follows.

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