Alternative definition of non-degenerate bilinear pairing. Let $V$ and $W$ be two vector spaces over field $\mathbb{k}$. Suppose $\beta : V \otimes W \rightarrow \mathbb{k}$ is a non-degenerate bilinear pairing in the variable $V$, then it means that for $0 \ne v \in V$, there exists an element $w \in W$ such that $v \otimes w \mapsto k$ where $k \ne 0$. This is the definition I remember (I hope I got it correct. Please correct me if I need to refine it).
However, recently I saw that there is another definition on non-degenerate bilinear pairing. It goes like this:
Let $V$ and $W$ be two vector spaces over field $\mathbb{k}$. A pairing $\beta : V \otimes W \rightarrow \mathbb{k}$ is called non-degenerate in the variable $V$ if there exists a linear map $\gamma : \mathbb{k} \rightarrow W \otimes V$, called co-pairing, such that the following composite is equal to the identity map of $V$:
\begin{equation*}
V = V\otimes \mathbb{k} \xrightarrow[]{id_V \otimes \gamma} V \otimes (W \otimes V) = (V \otimes W) \otimes V \xrightarrow[]{\beta \otimes id_V} \mathbb{k} \otimes V = V
\end{equation*}
Is this definition talking about the same non-degenerate bilinear pairing as the one I have in mind? If it is, then how are they equivalent? (They felt equivalent to me, but I don't see how)
 A: If $\gamma \colon \mathbb{k} \to W \otimes V$ is a  linear map then we dentote the composition
$$
    V
  = V \otimes \mathbb{k}
  \xrightarrow{\operatorname{id}_V \otimes \gamma} V \otimes W \otimes V
  \xrightarrow{\beta \otimes \operatorname{id}_V} \mathbb{k} \otimes V
  = V
$$
by $f_\gamma$. Notice that if $\gamma(1) = \sum_{i=1}^n w_i \otimes v_i$ then
$$
    f_\gamma(v)
  =  \sum_{i=1}^n \beta(v,w_i) v_i.
$$

Suppose that such a  a copairing exists, and let $v \in V$ with $v \neq 0$.
If $\gamma(1) = \sum_{i=1}^n w_i \otimes v_i$ then
$$
       0
  \neq v
  =    f_\gamma(v)
  =    \sum_{i=1}^n \beta(v,w_i) v_i,
$$
so at least one of the coeffients $\beta(v,w_i)$ must be nonzero.
Hence $\beta$ is non-degenerate in $V$.

For the existence we need $V$ to be finite-dimensional:
Notice that if $\gamma \colon \mathbb{k} \to W \otimes V$ is any linear map, then for $\gamma(1) = \sum_{i=1}^n w_i \otimes v_i$ we have by the above formula that
$$
    f_\gamma(v)
  = \sum_{i=1}^n \beta(v,w_i) v_i
  \in \langle v_1, \dotsc, v_n \rangle_{\mathbb{k}}
  \quad
  \text{for all $v \in V$}.
$$
Suppose that $\gamma$ is a copairing.
Then $f_\gamma(v) = v$ for all $v \in V$, and we get that $v \in \langle v_1, \dotsc, v_n \rangle_\mathbb{k}$ for all $v \in V$.
Thus $\langle v_1, \dotsc, v_n \rangle_\mathbb{k} = V$, i.e. $V$ is spanned by the finitely many vectors $v_1, \dotsc, v_n$, and is therefore finite-dimensional.
That shows that the existence of a copairing implies that $V$ is finite-dimensional.

So suppose that $V$ is finite-dimensional, say $\dim V =: n$, and that $\beta$ is non-degenerate in $V$.

Claim.
  The map $B \colon W \to V^*$, $w \mapsto \beta(-,w)$ is surjective.   
Proof.
  Suppose $B$ is not surjective.
  Then $\operatorname{im} B$ is a proper subspace of $V^*$, say $\dim \operatorname{im} B =: r < n$.
  We take a basis $(\phi_1, \dotsc, \phi_r)$ of $\operatorname{im} B$ and extend this to a basis $(\phi_1, \dotsc, \phi_n)$ of $V^*$.
  Let $(v_1, \dotsc, v_n)$ be the corresponding dual basis of $V$, i.e. for all $i,j = 1, \dotsc, n$ we have $\phi_i(v_j) = \delta_{ij}$.
  Then $\phi_i(v_n) = 0$ for all $i = 1, \dotsc, r$ and thus $\phi(v_n) = 0$ for all $\phi \in \operatorname{im} B$.
  With this we have $v_n \neq 0$ but
  $$
  \beta(v,w)
  = B(w)(v)
  = 0
  \quad
  \text{for all $w \in W$},
$$
  contradicting the non-degeneracy of $\beta$ in $V$.

Let $(v_1, \dotsc, v_n)$ be an ordered basis of $V$ and denote by $(v_1^*, \dotsc, v_n^*)$ the corresponding dual basis of $V^*$, i.e. $v_i^*(v_j) = \delta_{ij}$ for all $i,j = 1, \dotsc, n$.
Because $B$ is surjective there exists for every $i = 1, \dotsc, n$ some $w_i \in W$ with $v_i^* = B(w_i) = \beta(-,w_i)$.
Let $\gamma \colon \mathbb{k} \to W \otimes V$ be the unique linear map with $\gamma(1) = \sum_{i=1}^n w_i \otimes v_i$.
Then for all $j = 1, \dotsc, n$ we have
$$
    f_\gamma(v_j)
  = \sum_{i=1}^n \beta(v_j, w_i) v_i
  = \sum_{i=1}^n B(w_i)(v_j) v_i
  = \sum_{i=1}^n v_i^*(v_j) v_i
  = \sum_{i=1}^n \delta_{ij} v_i
  = v_j,
$$
and thus $f(v) = v$ for all $v \in V$ by the linearity of $f$.

So the existence of a copairing is equivalent to $V$ being finite-dimensional and $\beta$ being non-degenerate in $V$.
