Realtive Internal Hom Functor

Question

Let $(\mathfrak{C}, \otimes, I)$ be a pointed symmetric monoidal closed category and let $(M, \mu: M \otimes M \to M, \eta:I \to M)$ be an internal monoid in $\mathfrak{C}$. Assume we have an internal hom functor $[-,-]:\mathfrak{C}^{\operatorname{op}} \times \mathfrak{C} \to \mathfrak{C}$ such that for all $B \in \operatorname{Ob}(\mathfrak{C})$ there is an adunction $$ - \otimes B \dashv [B,-]:\mathfrak{C} \to \mathfrak{C}.$$ If $(A,\alpha:M \otimes A \to A)$ and $(B, \beta:M \otimes B \to B)$ are $M$-modules, what is the internal $M$-module hom $\operatorname{Hom}_{M}(A,B)$? I think that it should be realized as the equalizer between the $M$ action on $A$ and the $M$ action on $B$ on the internal hom object $[A,B]$, but I can't seem to figure out the correct morphisms to equalize.

Answer 1

Most of what you write are valid.

For a general monoid $M$, though, we don't get an $M$-action on $\hom_M(A,B)$, only if $M$ is commutative.

But we do get an internal hom object, as you say, as an equalizer of the two actions on the internal hom $[A,B]$.

By the defining adjunction, we have $$\matrix{X\otimes A\to B \\ \hline X\to [A,B]}$$ For the special case of $X=[A,B]$, we get the evaluation map $ev:[A,B]\otimes A\to B$ as the correspondent of $1_{[A,B]}$.

For the $A$-action, using that $\otimes$ is symmetric, we get an $[A,B]\to [M\otimes A,B]\cong [M,\,[A,B]]$ through the 'could-be-action' $M\otimes [A,B] \to [A,B]$, as $$\matrix{M\otimes A\otimes [A,B]\overset{\alpha}\to A\otimes [A,B] \overset{ev}\to B \\ \hline M\otimes [A,B] \to [A,B] \\ \hline [A,B] \to [M,\, [A,B]] }$$

For the $B$-action, it's $$\matrix{M\otimes [A,B]\otimes A\overset{ev}\to M\otimes B \overset{\beta}\to B \\ \hline M\otimes [A,B] \to [A,B] \\ \hline [A,B]\to [M,\,[A,B]]}\,.$$