Exponents in $\mathbf{Sets}^{G^{{\rm op}}}$ for an arbitrary group $G$.

Question

This is Exercise I.5(b) of Mac Lane and Moerdijk's, "Sheaves in Geometry and Logic [. . .]." According to Approach0, it is new to MSE.

The Details:

From p. 17 ibid. . . .

Definition 1: Given two functors

$$F:\mathbf{X}\to \mathbf{A}\quad G: \mathbf{A}\to \mathbf{X},$$

we say that $G$ is right adjoint to $F$, written $F\dashv G$, when for any $X\in{\rm Ob}(\mathbf{X})$ and any $A\in{\rm Ob}(\mathbf{A})$, there is a natural bijection between morphisms

$$\frac{X\stackrel{f}{\to}G(A)}{F(X)\stackrel{h}{\to}A},$$

in the sense that each $f$, as displayed, uniquely determines $h$, and conversely.

From p. 19 ibid. . . .

Definition 2: Suppose products exist in $\mathbf{C}$. For a fixed $A\in{\rm Ob}(\mathbf{C})$, one may consider the functor

$$A\times -: \mathbf{C}\to \mathbf{C}.$$

If this functor had a right adjoint (unique up to isomorphism), this adjoint is denoted by

$$(-)^A:\mathbf{C}\to \mathbf{C}.$$

In this case $A$ is said to be an exponentiable object of $\mathbf{C}$.

The Question:

For objects $X, Y$ in $\mathbf{Sets}^{G^{{\rm op}}}$, for $G$ a group, show that the exponent $Y^X$ can be described as the set of all functions $f: X\to Y$, with the right action of $g\in G$ on such a function defined by $(fg)x=[f(xg^{-1})]g$ for $x\in X$.

Thoughts:

I've answered Exercise I.5(a) with the help of Goldblatt's, "Topoi: A Categorial Analysis of Logic," since $\S$4.6(Exponentiation) defines exponentiation for left action by a monoid $M$.

Here is Exercise I.5(a):

In $\mathbf{B}M=\mathbf{Sets}^{M^{{\rm op}}}$ for $M$ a monoid observe that an object $X$ is a right action $X\times M\to X$ of $M$ on a set $X$ and that, $Y$ being another object, ${\rm Hom}(X, Y)$ is the set of equivariant maps $e:X\to Y$ [maps with $e(xm)=(ex)m$ for all $x\in X, m\in M$]. Prove that the exponent $Y^X$ is the set ${\rm Hom}(M\times X, Y)$ of equivariant maps $e: M\times X\to Y$, where $M$ is the set $M$ with right action by $M$, with the action $e\mapsto ek$ of $k\in M$ on $e$ defined by $(ek)(g, x)=e(kg, x)$.

Here is $\S$4.6(Exponentiation).

It's not too difficult to translate the description therein of the exponent.

Context:

For a rough idea of my abilities, see this question of mine. I am self-taught in category theory.

I think, given more time, that I should be able to do the exercise. But I've given it a couple of days and have got nowhere. I would like to move on to the next question.

Please help :)

Answer 1

Well you can try to prove that when $M$ is a group, then $\hom_{M-\mathbf{Set}}(M\times X,Y) \cong \hom_\mathbf{Set}(X,Y)$ and that under this isomorphism, the action given in I.5.(a) becomes the action described in the question (b) ?

Consider $$\begin{align} f: M\times X&\to M\times X,\\ (m,x)&\mapsto (m,xm^{-1}). \end{align}$$

Then $$\begin{align} f((m,x)\cdot_{M\times X} n) &= f(mn,xn) \\ &= (mn, xm^{-1}) \\ &= (m,xm^{-1})\cdot_{M\times X^{triv}} n\\ & = f(m,x)\cdot_{M\times X^{triv}} n, \end{align}$$ where $X^{triv}$ is $X$ with the trivial $M$-action.

It follows clearly that $M\times X\cong M\times X^{triv}$ as $M$-sets, so $\hom_{M-\mathbf{Set}}(M\times X,Y) \cong \hom_{M-\mathbf{Set}}(M\times X^{triv},Y) \cong \hom_\mathbf{Set}(X,Y)$ (the last isomorphism is easy to prove, and it's a good exercise)

Now you only have to see what the action on the left you described in (a) becomes on the right. It should be the action described in (b).