Let $k$ be a field and $V$ be a vector space over $k$ . Recall that $$V^{*}=\{f:V\to k\ |\ f\mbox{ is linear}\}$$ denotes the dual space of $V$ (which is also a vector space over $k$ ).

Proposition. Let $V$ and $W$ be vector spaces. Consider the map $\phi: V^{*}\otimes W^{*}\to (V\otimes W)^{*}$ such that $$\phi(f\otimes g)(v\otimes w)=f(v)g(w).$$ Then $\phi$ is a monomorphism. Moreover if one of the spaces $V$ , $W$ is finite dimensional, then $\phi$ is an isomorphism.

Proof. One can easily check that $\phi$ is a well defined linear map, thus it is sufficient to show that $\mathrm{Ker}(\phi)=0$ . So assume that $F\in V^{*}\otimes W^{*}$ is such that $\phi(F)=0$ . It is clear that $F$ can be (uniquely) expressed in the form $$F=\sum_{i,j} \alpha_{i,j} f_{i}\otimes g_{j},$$ where $(f_i)$ is a basis of $V^{*}$ , $(g_{j})$ is a basis of $W^{*}$ and $\alpha_{i,j}\in k$ . Then for any $v\in V$ and $w\in W$ we have: $$0=\phi(F)(v\otimes w)=\phi(\sum_{i,j}\alpha_{i,j} f_{i}\otimes g_{j})(v\otimes w)=$$ $$=\sum_{i,j}\alpha_{i,j} \phi(f_{i}\otimes g_{j})(v\otimes w)=\sum_{i,j}\alpha_{i,j} f_{i}(v) g_{j}(w).$$ Since $w\in W$ is arbitrary then we can write this equality in the form: $$0=\sum_{i,j}\alpha_{i,j}f_{i}(v)g_{j}=\sum_{j} (\sum_{i} \alpha_{i,j}f_{i}(v))g_{j}$$ and since $(g_{j})$ are linearly independent we obtain that $\sum_{i}\alpha_{i,j}f_{i}(v)=0$ for all $j$ . Again since $v\in V$ was arbitrary we obtain that $\sum_{i}\alpha_{i,j}f_{i}=0$ for all $j$ . Now since $(f_{i})$ are linearly independent we obtain that $\alpha_{i,j}=0$ for all $i,j$ . Thus $F=0$ .

Now assume that $\mathrm{dim}_{k}V=q<+\infty$ . Let $(v_i)_{i=1}^{q}$ be a basis of $V$ and let $(v_{i}^{*})_{i=1}^{q}$ be an induced basis of $V^{*}$ . Moreover let $(w_{p})_{p\,\in P}$ be a basis of $W$ . We wish to show that $\phi$ is onto, so let $f:V\otimes W\to k$ be an element of $(V\otimes W)^{*}$ . Define $F\in V^{*}\otimes W^{*}$ by the formula: $$F=\sum_{i=1}^{q} v_{i}^{*}\otimes g_{i},$$ where $g_{i}:W\to k$ is such that $g_{i}(w_{p})=f(v_i \otimes w_p)$ . Then for any $v_j$ from $(v_i)_{i=1}^{q}$ and for any $w_p$ from $(w_{p})_{p\,\in P}$ we have: $$\phi(F)(v_j\otimes w_p)=\phi(\sum_{i=1}^{q} v_{i}^{*}\otimes g_{i})(v_j\otimes w_p)=\sum_{i=1}^{q}\phi(v_{i}^{*}\otimes g_{i})(v_j \otimes w_p)=$$ $$=\sum_{i=1}^{q}v_{i}^{*}(v_{j})g_{i}(w_p)=g_{j}(w_p)=f(v_j\otimes w_p)$$ and thus $\phi(F)=f$ . $\square$

Remark. The map $\phi$ from the previous proposition is very important in studying algebras and coalgebras (more precisly it is an essence in defining dual (co)algebras). Unfortunetly $\phi$ does not have to be an isomorphism in general. Nevertheless, the spaces $(V\otimes W)^{*}$ and $V^{*}\otimes W^{*}$ are always isomorphic (see this entry for more details).