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 $V$.

Of course this depends on authors, but usually (especially when working over rings and not just fields), one says that a bilinear form is *nondegenerate* if it induces an *injection* from $V$ to its dual, and that it is *regular* when it induces an isomorphism.

For finite-dimensional vector spaces this is the same thing of course, but usually there is a difference, and in particular for infinite-dimensional vector spaces the notion of a regular bilinear form becomes void, as you noticed yourself, but you can still ask whether a form is nondegenerate.

## Best Answer

$$ (f:\text{not invertible} \iff\beta: \text{degenerate}) \equiv (f:\text{invertible} \iff\beta: \text{non-degenerate}) $$

## In the 1st argument

If $\beta$ is degenerate then there is a nonzero vector $v \in V$ such that for all $x \in V$ $$ 0 = \beta(v,x) = \langle v, f(x)\rangle \implies \text{Im}f \subset \{v\}^\perp \subset V, $$ so the image is proper subspace and $f$ is not surjective.

If $f$ is not invertible then its image is proper subspace $\text{Im} f \subsetneq V$, and wlog non trivial, then its orthogonal complement $W=(\text{Im} f)^\perp$ is nonempty. Take any $v \in W$ then for all $y \in V$ $$ \beta(v,y) := \langle v, f(y) \rangle = 0. $$

## In the 2nd Argument

If $\beta$ is degenerate then there is a nonzero vector $x \in V$ such that for all $w \in V$ $$ 0 = \beta(w,x) = \langle w, f(x)\rangle \implies f(x) = 0, $$

by the non-degeneracy of the inner product. It follows that the kernel is nontrivial and $f$ is not injective.Its kernel is not trivial and so there is a nonzero vector $y \in \ker f$ such that for all $v \in V$ the bilinear form $$ \beta(v,y) := \langle v, f(y) \rangle = \langle v, 0 \rangle = 0 $$ is degenerate.