Exercise 5

Let be a monoid considered as a category and consider . Objects in this category are sets equipped with a right action by . Maps are the equivariant maps with . When we refer to as an object in , we mean the set equipped with the trivial right action by from the monoid structure.

We wish to determine the exponent in this category.

Theorem: (a) The exponent in is the set itself equipped with the right action for each equivariant map and .

Proof: We define and show that it is right adjoint to . We will define the unit and co-unit and show that they satisfy the triangle identities.

First consider the unit , having components . Let assign to each a map by . To check that it is natural, consider an arrow . The naturality square is:

Take an element in the upper left. Following it down and right we obtain first the map and then the map . Following it right and down, we obtain and then . But of course these two expressions are the same precisely because is equivariant.

Now consider the counit (the evaluation map). It has components . For each pair let . The naturality square is:

Starting with a pair in the upper left, we follow it down and right to obtain first and then . Following it right and down we obtain first the pair and then .

Finally, we verify the triangle identities. First, we want to show the following is the identity:

Indeed, we have:

Next, we need to show that this is the identity:

Starting with a map , the first step obtains the map . The next step obtains the map . But by the right action on , we have . So the final map is just .

So the defined unit and counit are natural transformations satisfying the triangle identities, exhibiting an adjunction between and .

If the monoid is a group, the exponent is a bit “nicer”, in that it can be realized as the underlying exponent in with an appropriate group action.

Theorem: (b) In the category of right -sets, the exponent is the set equipped with the action for each equivariant map and .

Proof: Groups are, of course, monoids, and so we know that with the right-action . We exhibit an isomorphism between this -set and the one in the theorem. Simply define by . Define by . We verify that these are indeed equivariant maps that compose to the identity.

For , we need to verify that . On the left, we have:

On the other side,

where the penultimate step uses the fact that is an equivariant map.

For , we must again verify . On the left, we have

using the fact that is equivariant. On the right, we have

again using equivariance of .

Finally, observe that:


So , and we have exhibited an isomorphism between the two -sets.