# dual of tensor of vector spaces

Question regarding tensor products. Is this argument correct?

Let $b(V,\mathbb{R})$ be the vector space (in the natural way) of real valued bi-linear forms on the finite dimensional vector space $V$. By the universal property of the tensor product, each $b\in b(V,\mathbb{R})$ corresponds uniquely to a $\bar b\in (V\otimes V)^*$. I.e there exists $g$ such that $\bar b\circ g=b$, in fact $g(v,w)=v\otimes w$.

The map $b\rightarrow \bar b$ is an isomorphism since it is injective by the universal property and the dimension of the two spaces are the same.

Does this holds in general, i.e if $V$ is an arbitrary $R$-module and $b(V,R)$ denotes the $R-$module of $R$-bilinear maps from $V$ to $R$, and $(V\otimes V)^*$ is the $R$-module of $R$-linear maps from $V\otimes V$ to $R$ are these $R$-modules isomorphic? I.e how do one shows surjectivity?

One way to prove something is an isomorphism is to construct an inverse. By definition of the tensor product, for every $R$-bilinear map $b:V\times V\to R$ there is a unique homomorphism $\overline{b}:V\otimes_R V\to R$ such that composing $\overline{b}$ with the universal bilinear map $t:V\times V\to V\otimes_R V$ yields $b$, i. e. $b=\overline{b}\circ t$. Mapping $b\to \overline{b}$ gives a homomorphism from the module of bilinear maps $V\times V\to R$ into $\hom(V\otimes V,R)$. To construct an inverse, choose $c\in \hom(V\otimes V,R)$. Then we simply map $c$ to $c\circ t$, where again $t:V\times V\to V\otimes_R V$ is the universal bilinear map characterizing the tensor product. The fact that these constructions are inverses of each other follows from the universal property.