Given a field $k$ of null caracteristic, an associative/commutative/unital $k$-algebra $A$ and two free $k$-modules $M,N$, denote by $Hom\_k(M,A)$ the space of $k$-linear maps from $M$ to $A$.


I want to show that the following map is injective:


$Hom\_k(M,A)\otimes\_A Hom\_k(N,A) \longrightarrow Hom\_k(M \otimes\_k N,A)$


$ f \otimes\_A g \mapsto \Big( v \otimes\_k w \mapsto f(v)g(w)\Big)$


I might understand this in the finite dimensional case using the following string of isomorphisms


$Hom\_k(V,A) \otimes\_A Hom\_k(W,A) \simeq (V^\* \otimes\_kA)\otimes\_A(W^\*\otimes\_kA)\simeq V^\* \otimes\_k W^\* \otimes\_k A \simeq (V \otimes\_k W)^\* \otimes\_k A \simeq Hom\_k(V\otimes\_kW,A) $ 


But for some reason I can't figure it out in the general case.


Can someone give me some hint or a reference for it ?



Tensor of Hom spaces over an algebra
For infinite-dimensional algebras this is not true.


As it is mentioned in the comments, the claim is equivalent to the injectivity of the coordinate-wise product map $u:A^I \otimes\_A A^J \to A^{I\times J}$ (where now the exponent stands for products, not coproducts).


For a counterexample, consider the commutative, unital, associative algebra $A=\Bbbk [x\_i \; (i \in \mathbb{N})]/(x\_i x\_j \; (i,j \in \mathbb{N}))$, let $I=J=\mathbb{N}$ and $\alpha = (x\_i)\_{i \in \mathbb{N}} \in A^I$. Then clearly $u(\alpha \otimes \alpha)=0$, but one can prove that $\alpha \otimes \alpha$ is nonzero in $A^I \otimes\_A A^I$.


Indeed, $A^I \otimes\_A A^I = (A^I \otimes\_{\Bbbk} A^I)/U$, where $U$ is spanned by the elements of the form 
$$\lambda x\_i \otimes \mu - \lambda \otimes x\_i \mu,\quad
\lambda x\_i \otimes \beta,\quad
\beta \otimes x\_i \lambda$$
for all $i \in \mathbb{N}$, $\lambda, \mu\in \Bbbk^I$, $\beta \in N^I$ denoting $N=\mathrm{Span}(x\_k \mid k \in \mathbb{N})$. These elements do not span $\alpha \otimes \alpha$ in $A^I \otimes\_{\Bbbk} A^I$.


(**Edit**, response to the comment: If $A$ is finite-dimensional then $u$ is injective. Indeed, given a $\Bbbk$-basis $b\_1, \dots, b\_d$ in $A$, the elements of the form $\lambda \otimes \mu\, b\_k$ span $A^I \otimes\_A A^J$, where $k=1,\dots,d$, $\lambda \in \Bbbk^I$, $\mu \in \Bbbk^J$. Hence it is enough to prove the statement for $A=\Bbbk$.


Assume that $u$ vanishes on the nonzero element $\sum\_{k=1}^m \lambda^{(k)} \otimes \mu^{(k)}$ where $\lambda^{(k)},\mu^{(k)}$ are as above. We may assume that $\lambda^{(1)},\dots,\lambda^{(m)}$ are independent and similarly for $\mu$. Consequently, $\{(\lambda\_i^{(k)})\_k\mid i\in I\}$ spans $\Bbbk^m$ by a rank-argument.
On the other hand, as $u$ vanishes on the sum, we have 
$\sum\_{k=1}^m \lambda\_i^{(k)}\cdot \mu\_j^{(k)}=0$ for all $i,j$, i.e. $(\mu\_j^{(k)})\_k$ is orthogonal to a spanning set with respect to the obvious bilinear form on $\Bbbk^m$. This is a contradiction.)


