The standard definition of a (left) simplicial module $V$ over some simplicial algebra $A$ is the map of simplicial vector spaces $A\otimes V\to V$ that gives the usual modules component-wise. Here $\otimes$ denotes the component-wise tensor product of simplicial vector spaces (it gives to the category $sVect$ of simplicial vector spaces a symmetric monoidal structure).


But there is a natural structure of simplicial algebra on $sHom(V,V)$ (the internal $Hom$ induced by the product $\otimes$). So we can also define structure of a simplicial module over $A$ on $V$ by fixing a morphism of simplicial algebras $A\to sHom(V,V)$.


I think we can do it in more generality, for any symmetric monoidal category $C$ with internal $Hom$. If we have a monoid $A$ in $C$ and some object $V\in C$, then we can define the structure of $A$-module on $V$ in two different ways: via $A\otimes V\to V$ and $A\to \underline{Hom}(V,V)$, where $\underline{Hom}$ is the internal $Hom$.



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> The **question** is: are these two definitions equivalent for any symmetric monoidal category with internal $\underline{Hom}$? If not, is it true for simplicial vector spaces?
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Thank you very much!



Two definitions of modules in monoidal category
I will write $[B, C]$ instead of $\underline{\mathrm{Hom}}(B, C)$. Recall the tensor–hom adjunction:
$$\mathrm{Hom}(A \otimes B, C) \cong \mathrm{Hom}(A, [B, C])$$
Thus there is a canonical bijection between morphisms $\alpha : A \otimes V \to V$ and $\tilde{\alpha} : A \to [V, V]$. Let us show that $\alpha$ is an $A$-action on $V$ if and only if $\tilde{\alpha}$ is a monoid homomorphism. For simplicity I will work in a *strict* monoidal category.


* The unit axiom for $\alpha$ says, $\alpha \circ (e \otimes \mathrm{id}\_V) = \mathrm{id}\_V$; and the the unit axiom for $\tilde{\alpha}$ says, $\tilde{\alpha} \circ e = \eta\_V$, where $\eta\_V : I \to [V, V]$ is the right adjoint transpose of $\mathrm{id}\_V : V \to V$. The naturality of the tensor–hom adjunction in the first variable implies that these two conditions are equivalent.
* The compatibility axiom for $\alpha$ says, $\alpha \circ (m \otimes \mathrm{id}\_V) = \alpha \circ (\mathrm{id}\_A \otimes \alpha)$; and the compatibility axiom for $\tilde{\alpha}$ says, $\tilde{\alpha} \circ m = \mu\_V \circ (\tilde{\alpha} \otimes \tilde{\alpha})$, where $\mu\_V : [V, V] \otimes [V, V] \to [V, V]$ is the right adjoint transpose of the composite
$$[V, V] \otimes [V, V] \otimes V \xrightarrow{\mathrm{id}\_{[V, V]} \otimes \epsilon\_{V,V}} [V, V] \otimes V \xrightarrow{\epsilon\_{V,V}} V$$
where $\epsilon\_{B,C} : [B, C] \otimes B \to C$ is the left adjoint transpose of $\mathrm{id} : [B, C] \to [B, C]$. Now, naturality in the first variable implies the right adjoint transpose of $\alpha \circ (m \otimes \mathrm{id}\_V)$ is $\tilde{\alpha} \circ m$, and naturality in the first variable implies the left adjoint transpose of 
$\mu\_V \circ (\tilde{\alpha} \otimes \tilde{\alpha})$ is the following composite,
$$A \otimes A \otimes V \xrightarrow{\tilde{\alpha} \otimes \tilde{\alpha} \otimes \mathrm{id}\_V} [V, V] \otimes [V, V] \otimes V \xrightarrow{\mathrm{id}\_{[V, V]} \otimes \epsilon\_{V,V}} [V, V] \otimes V \xrightarrow{\epsilon\_{V,V}} V$$
but $\alpha = \epsilon\_{V,V} \circ (\tilde{\alpha} \otimes \mathrm{id}\_V)$, so the above reduces to $\alpha \circ (\mathrm{id}\_A \otimes \alpha)$.


Thus the claim is proved.


