Why is $V^{\vee}\otimes W^{\vee}\longrightarrow (V\otimes W)^{\vee}$ always injective?

Let $$R$$ be a commutative ring with $$1$$. For all $$R$$-modules $$V,W$$ we have a canonical $$R$$-linear map $$V^{\vee}\otimes W^{\vee}\longrightarrow (V\otimes W)^{\vee}$$ from tensor product of dual modules into the dual of tensor product. My question is, why is this map always injective?

This is false. For instance, let $$k$$ be a field and let $$R=k[x_0,x_1,x_2,\dots]/(x_0,x_1,x_2,\dots)^2$$. Let $$V=W$$ be a free $$R$$-module with basis $$\{e_0,e_1,\dots\}$$ and let $$\alpha\in V^\vee$$ be given by $$\alpha(e_i)=x_i$$. Note that $$\alpha\otimes\alpha$$ maps to $$0$$ in $$(V\otimes_R V)^\vee$$, since $$\alpha(e_i)\alpha(e_j)=x_ix_j=0$$ for any $$i,j$$. However, I claim $$\alpha\otimes\alpha$$ is nonzero in $$V^\vee\otimes_R V^\vee$$.
To prove this, note that $$R$$ is the direct limit of the subrings $$R_n=k[x_0,\dots,x_n]/(x_0,\dots,x_n)^2$$ and so $$V^\vee\otimes_R V^\vee$$ is the direct limit of the tensor products $$T_n=V^\vee\otimes_{R_n} V^\vee$$ over $$R_n$$. It thus suffices to show that $$\alpha\otimes\alpha$$ is nonzero in $$T_n$$ for all $$n$$.
Note that $$V^\vee\cong R^\mathbb{N}$$, with the coordinates corresponding to evaluation at the $$e_i$$. As an $$R_n$$-module, $$R$$ splits as a direct sum $$R_n\oplus k^{\oplus\mathbb{N}}$$ where $$k$$ is an $$R_n$$-module by letting the variables act trivially and the second summand is the vector space spanned by the $$x_i$$ for $$i>n$$. Thus as an $$R_n$$-module, $$V^\vee\cong R_n^{\mathbb{N}}\oplus (k^{\oplus\mathbb{N}})^{\mathbb{N}}$$; let us call these two summands $$M$$ and $$N$$. Now notice that $$N$$ is annihilated by the maximal ideal of $$R_n$$ and so $$N\otimes_{R_n} N$$ can be identified with $$N\otimes_k N$$. But when you take a tensor product over a field, the tensor of any two nonzero vectors is nonzero. In particular, since the projection of $$\alpha$$ onto $$N$$ is nonzero, so is the projection of $$\alpha\otimes \alpha\in T_n$$ onto $$N\otimes_{R_n} N$$. Thus $$\alpha\otimes\alpha$$ is nonzero in $$T_n$$, as desired.