Another (covariant) functor

I have one more example of a (covariant) functor that I thought is worth mentioning. It’s so important, it has it’s own name.

Let \mathcal C be a locally small category. Then for each X,Y\in Ob(\mathcal C), \hom_{\mathcal C}(X,Y) is a set. Pick a special object S in \mathcal C, and define a functor h_S:\mathcal C\to\textsc{Set} in the following way:

If X\in Ob(\mathcal C), h_S:X\mapsto \hom(S,X). We need \mathcal C to be locally small to ensure that \hom(S,X) really is a set. For a morphism f:X\to Y in h_S(f) is a map from \hom(S,X) to \hom(S,Y). It sends \phi:S\to X to f\circ\phi:S\to Y. That is,

h_S(f):\phi\mapsto f\circ\phi.

We call these functors representable functors. Yes, this is complicated. The objects aren’t so hard, but the morphisms are confusing. I don’t know a good way to understand them. Beyond working through the definitions and checking that this really is a functor, I don’t know any good exercises. Try to do it yourself. If you get stuck, you can look below at what I’ve done, but as is almost always true in math, it’s best to do it yourself if you can.


We consider h_S(1_X):\hom(S,X)\to\hom(S,X) by \phi\mapsto 1_X\circ\phi=\phi. As a set map, this is simply the identity function on the set \hom(S,X). That is, it is 1_{\hom(S,X)}. Another way to write this is 1_{h_S(X)}.

Now let f:X\to Y and g:Y\to Z be morphisms in \mathcal C. This gives us h_S(g\circ f):\phi\mapsto (g\circ f)\circ\phi. We can break this map into two pieces. First, we do the piece that composes with f. Then we do the part that composes with g. That is, first we have h_S(f):\phi\mapsto f\circ\phi, then we apply the map h_S(g) to get h_S(g)\circ h_S(f):\phi\mapsto g\circ f\circ \phi. That is,

h_S(f)\circ h_S(g)=h_S(f\circ g)

as desired. Another way to check this is to check that the diagram

commutes. Let’s check it again thinking about it this way. To see if a diagram commutes, we need to check that all paths give us the same result. In this case, there’s only one thing to check. Take \phi\in\hom(S,X).

One way to get from \hom(S,X) to \hom(S,Z) gives us h_S(g\circ f)(\phi)=(g\circ f)\circ \phi. The other way is by composing h_S(f) and h_S(g). If we start with \phi, and apply the map h_S(f), we get f\circ \phi. After applying the map h_S(g), we get g\circ(f\circ \phi). By associativity, these are the same results, which means the diagram commutes.

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One Response to Another (covariant) functor

  1. Pingback: Yoneda’s Lemma (part 1) « Andy Soffer

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