Why is the category of $\textit{right } A$-modules a $\textit{left } \mathcal{C}$-module category? Let $C$ be a monoidal category and let $A$ be an algebra in $C$. 
Why is the category $Mod_C(A)$ of right $A$-modules a left $C$-module category?
Take any $Y \in Mod_C(A)$ and any $X \in C$. We want to show that we have a tensor product $C \times Mod_C(A) \rightarrow Mod_C(A)$, so we need to prove that $XY \in Mod_C(A)$.
We know that there is a $m: YA \rightarrow Y$, but I'm not sure how we can use that fact to show the above. Am I missing something? Why isn't $Mod_C(A)$ a right $C$-module category?
 A: It is very simple. If $X \in C$ is any object and $(Y, m : Y \otimes A \to Y)$ is a right $A$-module, then $(X \otimes Y, X \otimes m : X \otimes Y \otimes A \to X \otimes Y)$ is a right $A$-module (one easily checks the module axioms). This defines a functor $C \times \mathrm{Mod}_C(A) \to \mathrm{Mod}_C(A)$.
Notice that in general there is no right $A$-module structure on $Y \otimes X$, since we cannot do anything with $Y \otimes X \otimes A$ (when $C$ is not symmetric).
It might be a bit counterintuitive that right modules produce a left module category. But this is actually a general theme. For example, let $C=\mathbf{Set}$ (or $\mathbf{Ring}$), so $A$ is a monoid (or a ring), and consider $A$ as a right $A$-module. Then $\mathrm{End}_{\mathrm{Mod}(A)}(A)$ is isomorphic to $A$ (not $A^{op}$), which implies that $A$ is actually a left $A$-module object in $\mathrm{Mod}(A)$. And actually, it is the universal cocomplete category with a left $A$-module object. More generally, if $\mathcal{L}$ is any Lawvere theory, then $\mathrm{Mod}(L)$ is the universal cocomplete category with an $\mathcal{L}^{op}$-model.
Here is a more elementary manifestation of this theme: If $R$ is a ring and $I \subseteq R$ is a right ideal, then for every $r \in R$ the left multiplication $r \cdot I$ is a right ideal of $R$ as well. This is because the left multiplication $x \mapsto r \cdot x$ is a right $R$-module homomorphism.
