Let $M$ be a monoid considered as a category and consider $\mathbf{B}M = \presheaves{M}$. Objects in this category are sets equipped with a right action by $M$. Maps $f : X \rightarrow Y$ are the equivariant maps $X \rightarrow Y$ with $f(xm) = f(x)m$. When we refer to $M$ as an object in $\mathbf{B}M$, we mean the set $M$ equipped with the trivial right action by $M$ from the monoid structure.

We wish to determine the exponent in this category.

Theorem: (a) The exponent $Y^X$ in $\mathbf{B}M$ is the set $\Hom(M \times X, Y)$ itself equipped with the right action $(km)(n, x) = k(mn, x)$ for each equivariant map $k : M \times X \rightarrow Y$ and $m \in M$.

Proof: We define $(\dash)^X = \Hom(M \times X, \dash)$ and show that it is right adjoint to $\dash \times X$. We will define the unit and co-unit and show that they satisfy the triangle identities.

First consider the unit $\eta : 1_{\mathbf{B}M} \rightarrow (\dash \times X)^X$, having components $\eta_Y : Y \rightarrow (Y \times X)^X$. Let $\eta_Y$ assign to each $y \in Y$ a map $M \times X \rightarrow Y \times X$ by $\eta_Y(y)(m, x) = (ym, x)$. To check that it is natural, consider an arrow $f : Y \rightarrow Z$. The naturality square is:

Take an element $y$ in the upper left. Following it down and right we obtain first the map $(m, x) \mapsto (ym, x)$ and then the map $(m, x) \mapsto (f(ym), x)$. Following it right and down, we obtain $f(y)$ and then $(m, x) \mapsto (f(y)m, x)$. But of course these two expressions are the same precisely because $f$ is equivariant.

Now consider the counit $\varepsilon : (\dash)^X \times X \rightarrow 1_{\mathbf{B}M}$ (the evaluation map). It has components $\varepsilon_Y : Y^X \times X \rightarrow Y$. For each pair $k : M \times X \rightarrow Y, x \in X$ let $\varepsilon_Y(k, x) = k(1_M, x)$. The naturality square is:

Starting with a pair $(k, x)$ in the upper left, we follow it down and right to obtain first $k(1_M, x)$ and then $f(k(1_M, x))$. Following it right and down we obtain first the pair $(f \circ k, x)$ and then $(f \circ k)(1_M, x) = f(k(1_M, x))$.

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

$$Y \times X \xrightarrow{\eta_Y \times 1_X} (Y \times X)^X \times X \xrightarrow{\varepsilon_{Y \times X}} Y \times X$$

Indeed, we have:

$$(y, x) \mapsto ((m, x) \mapsto (ym, x), x) \mapsto (y 1_M, x) = (y, x)$$

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

$$Y^X \xrightarrow{\eta_{\left(Y^X\right)}} (Y^X \times X)^X \xrightarrow{\varepsilon_Y \circ \dash} Y^X$$

Starting with a map $k : M \times X \rightarrow X$, the first step obtains the map $(m, x) \mapsto (km, x)$. The next step obtains the map $(m, x) \mapsto (km)(1_M, x)$. But by the right action on $\Hom(M \times X, Y)$, we have $(km)(1_M, x) = k(m1_M, x) = k(m, x)$. So the final map is just $k$.

So the defined unit and counit are natural transformations satisfying the triangle identities, exhibiting an adjunction between $\dash \times X$ and $(\dash)^X$.

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If the monoid is a group, the exponent is a bit "nicer", in that it can be realized as the underlying exponent in $\mathbf{Set}$ with an appropriate group action.

Theorem: (b) In the category $\mathbf{B}G$ of right $G$-sets, the exponent $Y^X$ is the set $\Hom(X, Y)$ equipped with the action $(fp)(x) = f(xp^{-1}) p$ for each equivariant map $f : X \rightarrow Y$ and $p \in G$.

Proof: Groups are, of course, monoids, and so we know that $Y^X = \Hom(G \times X, Y)$ with the right-action $(kp)(q, x) = k(pq, x)$. We exhibit an isomorphism between this $G$-set and the one in the theorem. Simply define $\varphi : \Hom(G \times X, Y) \rightarrow \Hom(X, Y)$ by $\varphi(k)(x) = k(1_G, x)$. Define $\phi : \Hom(X, Y) \rightarrow \Hom(G \times X, Y)$ by $\phi(f)(q, x) = f(xq^{-1})q$. We verify that these are indeed equivariant maps that compose to the identity.

For $\varphi$, we need to verify that $\varphi(kp) = \varphi(k)p$. On the left, we have:

$$\varphi(kp)(x) = (kp)(1_G, x) = k(p 1_G, x) = k(p, x)$$

On the other side,

$$\left[ \varphi(k) p \right](x) = \left[ \varphi(k)(xp^{-1}) \right] p = k(1_G, xp^{-1})p = k(1_G p, x p^{-1} p) = k(p, x)$$

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

For $\phi$, we must again verify $\phi(fp) = \phi(f)p$. On the left, we have

$$\phi(fp)(q, x) = \left[ (fp)(xq^{-1}) \right]q = f(xq^{-1}p^{-1})p q = f(xq^{-1}p^{-1}pq) = f(x)$$

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

$$\left[ \phi(f)p \right](q, x) = \phi(f)(pq, x) = f(x(pq)^{-1})pq = f(x(pq)^{-1}pq) = f(x)$$

again using equivariance of $f$.

Finally, observe that:

$$\varphi(\phi(f))(x) = \phi(f)(1_G, x) = f(x1_G)1_G^{-1} = f(x)$$

and

$$\phi(\varphi(k))(q, x) = \left[ \varphi(k)(xq^{-1}) \right]q = k(1_G, xq^{-1})q = k(1_G q, xq^{-1}q) = k(q, x)$$

So $\varphi \circ \phi = \phi \circ \varphi = 1$, and we have exhibited an isomorphism between the two $G$-sets.