Can the canonical map from a tensor product of Hom's to the Hom of tensor products fail to be injective?

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Is there an example of a commutative ring $K$ and modules $V_1, V_2, W_1, W_2$ such that the canonical linear map $\operatorname{Hom}(V_1, W_1) \otimes \operatorname{Hom}(V_2, W_2) \to \operatorname{Hom}(V_1 \otimes V_2, W_1 \otimes W_2)$ fails to be injective?

Certainly it is always injective in the case where these are all free, and it is bijective in the case of finitely generated free modules.

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Yes, there are many such examples. A neat way to find some is to manage to get $W_1\otimes W_2 = 0$ without having the LHS vanish.

One such example is given by $ W_1=W_2 = \mathbb{Q/Z}$. Then $W_1\otimes W_2 = 0$ so the RHS is $0$, so it suffices to find $V_1,V_2$ such that the LHS is nonzero, and then the map will automatically fail to be injective.

Take, e.g. $V_1=V_2 = \mathbb Z/n$ with $n>0$, then $\hom(V_1,W_1) \cong \mathbb Z/n$ (the $n$-torsion of $\mathbb{Q/Z}$), and so the LHS is $\mathbb Z/n \otimes \mathbb Z/n $ which is always nonzero for $n\geq 2$.