Let $k$ be a field, $V,W$ be two vector spaces over $k$ and $\beta : V \times W \to k$ be a bilinear form.
I have always seen non-degeneracy for $\beta$ over $V$ stated in the following way : $$\forall \,v \in V\setminus\{0\}\,\,,\,\, \exists w \in W \,\,,\,\, \beta(v,w)\neq 0. $$
While reading Kock's book "Frobenius algebras and 2D topological quantum field theories" I read the following definition.
If we write $\bar\beta : V \otimes W \to k$ for the corresponding linear map then $\bar\beta$ is non-degenerate over $V$ if there exists some linear map $\gamma : k \to W \otimes V$ such that the map $$ V = V\otimes k \xrightarrow{\text{$id_V\otimes\gamma$}} V \otimes W \otimes V \xrightarrow{\beta\otimes id_V} k \otimes V = V $$ is the identity of $V$.
My question is : are this two definitions equivalent ?
I have already shown that the second implies the first but I'm completely stuck when I try to do the converse.
No, they are not equivalent. The second definition is equivalent to the condition that the map $V \ni v \mapsto \beta(v, -) \in W^{\ast}$ is an isomorphism (equivalently, that the map $W \ni w \mapsto \beta(-, w) \in V^{\ast}$ is an isomorphism). The first definition only states that the first map is injective; in particular, it does not imply that $V, W$ are finite-dimensional.