= Solution
We use the left internal-hom convention in which $[M,N]$ is right adjoint to $-\otimes_kM$, so its <evaluation morphism> has domain $[M,N]\otimes_kM$. For a <Hopf algebra> over the <commutative ring> $k$, take the full $k$-module
$$
[M,N]=\operatorname{Hom}_k(M,N)
$$
and give it the <internal hom for modules over a Hopf algebra>
$$
\boxed{(h\cdot f)(u)=\sum h_{(1)}\cdot f(S(h_{(2)})\cdot u).}
$$
The <antipode> is an antihomomorphism, $S(hg)=S(g)S(h)$, so
$$
(h\cdot(g\cdot f))(u)=\sum h_{(1)}g_{(1)}\cdot f(S(g_{(2)})S(h_{(2)})\cdot u)=((hg)\cdot f)(u).
$$
The unit acts identically. Thus this is a left $H$-<module>.
The <evaluation morphism> is $H$-linear. Using the <diagonal bialgebra action> and the antipode identity,
$$
\sum(h_{(1)}\cdot f)(h_{(2)}\cdot u)=\sum h_{(1)}\cdot f(S(h_{(2)})h_{(3)}\cdot u)=h\cdot f(u).
$$
Now let $q:L\otimes_kM\to N$ be $H$-linear and define its ordinary curried map $\widehat q(\ell)(u)=q(\ell\otimes u)$. To see that it too is $H$-linear, compute
$$
\begin{aligned}
(h\cdot\widehat q(\ell))(u)&=\sum h_{(1)}\cdot q(\ell\otimes S(h_{(2)})\cdot u)\\
&=\sum q(h_{(1)}\cdot\ell\otimes h_{(2)}S(h_{(3)})\cdot u)\\
&=q(h\cdot\ell\otimes u)=\widehat q(h\cdot\ell)(u).
\end{aligned}
$$
Conversely, any $H$-linear map $L\to[M,N]$ uncurries to an $H$-linear map by the already established linearity of evaluation. The ordinary tensor–hom <adjunction> therefore restricts to a natural <bijection>
$$
\boxed{\operatorname{Hom}_H(L\otimes_kM,N)\cong\operatorname{Hom}_H(L,[M,N]).}
$$
Postcomposition by an $H$-linear map makes $[M,-]$ a <functor>. This proves that the category of left $H$-<modules> is a <left closed monoidal category>. \b[All modules are allowed; no inverse antipode or finite-dimensional dual is required.] Stating the tensor–hom convention explicitly avoids confusing this construction with the closure on the opposite side.
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