$\bf M-Set$ is cartesian closed

Question

I must show that, given a monoid $M$, the category $\bf \text{M-Set}$ is cartesian closed. Take a $M\text{-Set}$ $X$, and call $-^X$ the right adjoint of $-\times X$. For any $M\text{-Set}$ $Y$, $Y^X$ must be a $M\text{-Set}$; in first place we'd like to describe its underlying set. This can be done with the following natural bijections: $$U(Y^X)\cong\mathbf{Set}(*,U(Y^X))\cong\mathbf{ \text{M-Set}}(M,Y^X)\cong \mathbf {\text{M-Set}}(M\times X,Y).$$ Now I should equip $\mathbf {\text{M-Set}}(M\times X,Y)$ with an action of $M$ such that it becomes our right adjoint; this means that there must be a $M$-homomorphism $\eta _Y:Y\to \mathbf {\text{M-Set}}(M\times X,Y\times X)$ such that, for every $M\text{-Set}$ $Z$ and $M$-homomorphism $f:Y\to\mathbf {\text{M-Set}}(M\times X,Z)$, there is a unique $M$-homomorphism $\bar f:Y\times X\to Z$ such that $f=\mathbf {\text{M-Set}}(M\times X,\bar f)\circ \eta_Y$. Now, from this point of view it seems quite complex to me to see the solution; however my teacher left this as an exercise so it must be a simple problem. Can you suggest me how to proceed?

Answer 1

Fix $y\in Y$. Given $(m,x)\in M\times X$, what possible ways you have to get an element of $Y\times X$? I can think of $(y,x)$, $(my,x)$, $(y,mx)$ and $(my,mx)$. Three of these don't define a $M$-morphism $M\times X\to Y\times X$, so you already have what $\eta_Y(y)$ should be.

After you've done this, you can reverse engineer what the action on $(Y\times X)^X$ must be in order for $\eta_Y$ to be an $M$-morphism. The formula you will get won't depend on the codomain, so you actually got the action on any $Z^X$.