Injection vs. bijection between $V^\star \otimes V^\star$ and $(V \otimes V)^\star$ [closed]

Question

It is a well-known fact that, if $V$ is a vector space over a field $k$, then $V^\star \otimes V^\star$ embeds into $(V \otimes V)^\star$.

It turns out to be an isomorphism when $V$ is a finite-dimensional vector space.

Why is this finite-dimensional restriction necessary?

More explicitly, where is the mistake in the following reasoning?

Let $\mathcal{B}$ be (an infinite) basis of $V$. For all $b_0 \in \mathcal{B}$, we consider the map $b_0^\star : V \longrightarrow k$ defined for all $b \in \mathcal{B}$ by $$ b_0^\star(b) = \left \{ \begin{array}{ll} 1 & \text{, if }b = b_0\\ 0 & \text{, otherwise} \end{array} \right. $$ Consequently, $\mathcal{B}^\star = \{b^\star, b \in \mathcal{B}\}$ is a basis of $V^\star$. (Does $\mathcal{B}^\star$ really span $V^\star$?)

Similarly, we also consider for all $b_{01}, b_{02} \in \mathcal{B}$ the map $(b_{01} \otimes b_{02})^\star : V \otimes V \longrightarrow k$ defined for all $b_1, b_2 \in \mathcal{B}$ by $$ (b_{01} \otimes b_{02})^\star(b_1 \otimes b_2) = \left \{ \begin{array}{ll} 1 & \text{, if }b_{01} = b_1 \text{ and } b_{02} = b_2\\ 0 & \text{, otherwise} \end{array} \right. $$ Consequently, $\mathcal{C}^\star = \{(b_1 \otimes b_2)^\star, b_1, b_2 \in \mathcal{B}\}$ is a basis of $(V \otimes V)^\star$. (Does $\mathcal{C}^\star$ really span $(V \otimes V)^\star$?)

Consequently, the map $\varphi: V^\star \otimes V^\star \longmapsto (V \otimes V)^\star$ defined by $\varphi(b_1^\star \otimes b_2^\star) = (b_1 \otimes b_2)^\star$ for all $b_1, b_2 \in \mathcal{B}$ is well-defined and bijective, with an inverse bijection defined by $\varphi^{-1}\big ( (b_1 \otimes b_2)^\star \big ) = b_1^\star \otimes b_2^\star$.

Finally, do we know an explicit example showing that $V^\star \otimes V^\star \hookrightarrow (V \otimes V)^\star$ is not an isomorphism when $V$ is an infinite-dimensional vector space?

Answer 1

Write $V=k^{(I)}$ (this is always the case, up to isomorphism, for some set $I$, namely take $I$ to be a basis subset).

So $V^*=K^I$, $V\otimes V=k^{(I\times I)}$ and the embedding of $V^I\otimes V^I$ into $V^{I\times I}$ in consideration maps $f\otimes g$ to $(f\otimes g)(x,y)=f(x)g(y)$.

I claim that the Kronecker function $(x,y)\mapsto\delta_{x,y}$ is not in the image.

Otherwise, we can write it as $(x,y)\mapsto \sum_{i=1}^nf_i(x)g_i(y)$. We choose $n$ minimal (over all infinite $I$).

So $f_n\neq 0$. Fix $x_0$ with $f_n(x_0)\neq 0$. Then for all $y\neq 0$ we have $\sum_{i=1}^n f_i(x_0)g_i(y)=0$. Hence on $I\smallsetminus\{x_0\}$ we have $g_n=-\sum_{i=1}^{n-1}(f_i(x_0)/f_n(x_0))g_i$. On $I'\times I'$ we rewrite the condition $\sum_{i=1}^n f_i(x)g_i(y)=\delta_{x,y}$ by replacing $g_n$ with the above linear combination. Then we obtain a similar decomposition of $\delta_{\cdot,\cdot}$ with $n$ replaced with $n-1$. We thus contradict the minimality of $n$.