How to prove that the canonical inclusion of $V_1^\ast\otimes \cdots \otimes V_n^\ast$ into $(V_1\otimes\cdots\otimes V_n)^\ast$ is an inclusion

Question

I read these threads: 1, 2, 3, but my question is somewhat simpler, maybe trivial — sorry in advance, if this is the case.
How do we prove, in the general case, that the canonical inclusion is indeed an inclusion?
By using a couple of times the universal property of the tensor product (on different environments) we build a linear map, let's say $F: V_1^\ast\otimes\dots\otimes V_n^\ast \to(V_1\otimes\dots\otimes V_n)^\ast$, such that:

$$F(\phi_1\otimes\dots\otimes\phi_n)\left(\boldsymbol{v}_1\otimes \dots\otimes\boldsymbol{v}_n\right)=\phi_1(\boldsymbol{v}_1)\dots\phi_n(\boldsymbol{v}_n);$$

for any $(\phi_1,\dots,\phi_n)\in V_1^\ast\times\dots\times V_n^\ast$ and $(\boldsymbol{v}_1, \dots,\boldsymbol{v}_n)\in V_1\times\dots\times V_n$; so… how do we get that $F$ is injective?
I'm ok with the finite-dimensional case, but I can't see how to adapt its proof to the general case: the one I'm aware of rests upon the fact that each linear functional (covector) can be decomposed on the dual basis.
Of course I tried to show that the kernel of $F$ is trivial; now, I can see why — hope what I'm about to write is not wrong — if $F$ vanishes on a simple tensor $\phi_1\otimes\dots\otimes\phi_n$, then this latter vanishes too; but then $V_1^\ast\otimes\dots\otimes V_n^\ast$ is made up of all the finite sums of such tensors, so how do I manage such a case?

I'm not familiar with categories, so, in order to be sure that I can follow your answer, please keep things as simple as possible; eg, should pairings, evaluation contractions, and/or canonical isomorphisms be needed, please don't give for granted that I'm already able to follow you on everything.

Answer 1

I will show it in the case of two factors $$V^{\star} \otimes W ^{\star} \to (V \otimes W)^{\star}$$

Consider an element on LHS $\sum \alpha_{ij} k_i\otimes l_j$ where $k_i \in V_1^{\star}$, $l_j \in V_2^{\star}$ linearly independent. We know that for every $v \otimes w \in V_1 \otimes V_2$ we have $$\sum \alpha_{ij}l_i(v) k_j (w)=0$$

Now fix $w \in W$. Then we have for every $v\in V$ $$\sum_i( \sum \alpha_{ij} k_j(w)) l_i(v) = 0$$ Since the $l_i$ are linearly independent we get $$\sum_j \alpha_{ij}k_j(w)=0$$ for all $i$. Now since the $k_j$ are also independent we get $\alpha_{ij} =0$ for all $i$, $j$.

Note that we did not use anywhere the fact that the $l_i$, $k_j$ are linear. This is connected to the fact that the map
$$\mathcal{F}(X, \mathbb{K}) \otimes \mathcal{F}(Y, \mathbb{K})\to \mathcal{F}(X\times Y, \mathbb{K})$$ is injective, where $\mathcal{F}(X, \mathbb{K})$ is a space of functions.

Answer 2

I will prove this in the case $n=2$; the argument for general $n$ is almost identical but the notation is messier.

So, suppose $V$ and $W$ are vector spaces over a field $k$, and we wish to show your map $F:V^*\otimes W^*\to (V\otimes W)^*$ is injective. The idea is that every element of $V^*\otimes W^*$ actually comes from some finite-dimensional quotients of $V$ and $W$, so that you can reduce to the case where $V$ and $W$ are finite-dimensional.

Specifically, suppose $x=\sum_{i=1}^m f_i\otimes g_i\in \ker(F)$. Let $f:V\to k^m$ and $g:W\to k^m$ be the linear maps whose coordinates are the $f_i$ and $g_i$. Let $V_0=\ker f$ and $W_0=\ker g$. Note that the quotient map $V\to V/V_0$ induces a linear injection $(V/V_0)^*\to V^*$ (the inclusion of the functionals on $V$ that happen to vanish on $V_0$), and similarly there is a linear injection $(W/W_0)^*\to W^*$. Tensoring these together we get a linear injection $\alpha:(V/V_0)^*\otimes (W/W_0)^*\to V^*\otimes W^*$. By construction, our element $x=\sum f_i\otimes g_i$ is in the image of $\alpha$, since each $f_i$ vanishes on $V_0$ and each $g_i$ vanishes on $W_0$. Also, $V/V_0$ and $W/W_0$ are finite-dimensional, since they are isomorphic to the images of $f$ and $g$.

We can also tensor the quotient maps together to get a linear surjection $V\otimes W\to V/V_0\otimes W/W_0$ which dualizes to a linear injection $\beta:(V/V_0\otimes W/W_0)^*\to (V\otimes W)^*$. These maps all fit into a commutative diagram $$\require{AMScd} \begin{CD} (V/V_0)^*\otimes (W/W_0)^* @>{F'}>> (V/V_0\otimes W/W_0)^*\\ @V{\alpha}VV @V{\beta}VV \\ V^*\otimes W^* @>{F}>> (V\otimes W)^* \end{CD}$$ where the top map $F'$ is just your $F$ for the pair of vector spaces $V/V_0$ and $W/W_0$ instead of $V$ and $W$. Since $V/V_0$ and $W/W_0$ are finite-dimensional, we know that $F'$ is an isomorphism. Now, our element $x\in \ker(F)$ is in the image of $\alpha$; say $x=\alpha(y)$. We then have $$0=F(x)=F(\alpha(y))=\beta(F'(y)).$$ But $\beta$ and $F'$ are injective, so this implies $y=0$, so $x=\alpha(y)=0$. Thus we have shown an arbitrary element of $\ker F$ is $0$, so $F$ is injective.