TITLE: Is $(V_1\otimes\cdots\otimes V_k)^\ast \simeq V_1^\ast\otimes \cdots \otimes V_k^\ast$ true for infinite dimensional spaces?
QUESTION [13 upvotes]: Suppose $V_1,\dots,V_k$ are vector spaces of finite dimension. Then I could prove easily that $(V_1\otimes\cdots\otimes V_k)^\ast\simeq V_1^\ast\otimes\cdots\otimes V_k^\ast$. My proof was like that: first of all, I've shown that if for each $i$ we have $W_i$ another vector space such that $V_i\simeq W_i$ then $V_1\otimes\cdots\otimes V_k \simeq W_1\otimes\cdots\otimes W_k$.
Then, since I'm supposing each $V_i$ finite dimensional, each $V_i\simeq V_i^\ast$ and also, we have $V_1\otimes\cdots\otimes V_k$ also finite dimensional, so that
$$(V_1\otimes\cdots\otimes V_k)^\ast \simeq V_1\otimes\cdots\otimes V_k\simeq V_1^\ast\otimes \cdots \otimes V_k^\ast$$
and so it is proved. Now, if the spaces are not finite dimensional this proof cannot be used. In that case, the property still holds? Is it possible to prove for infinite dimensional spaces?
I've tried to prove it directly, constructing an isomorphism. I've picked first the mapping $\psi : V_1^\ast\times\cdots\times V_k^\ast \to \mathcal{L}(V_1,\dots,V_k;\mathbb{K})$ given by
$$\psi(f_1,\dots,f_k)(v_1,\dots,v_k) = f_1(v_1)\cdots f_k(v_k)$$
this map is multilinear and hence by the universal property there corresponds a unique linear mapping $\phi : V_1^\ast\otimes \cdots \otimes V_k^\ast \to \mathcal{L}(V_1,\dots,V_k;\mathbb{K})$ such that:
$$\phi(f_1\otimes\cdots\otimes f_k)(v_1,\dots,v_k) = f_1(v_1)\cdots f_k(v_k)$$
To show that this $\phi$ is isomorphis I would need to find an inverse, but I didn't have any idea. Is it possible to complete this proof?
Thanks very much in advance.

REPLY [5 votes]: The situation here is a little subtle.
The canonical inclusion $(V^*_1\otimes\dots\otimes V_k^*)\hookrightarrow(V_1\otimes\dots\otimes V_k)^*$ isn't an isomorphism (it's a strict injection), but (assuming the axiom of choice) the spaces are nevertheless isomorphic by some other (non-canonical) map.
The proof that the canonical map isn't an isomorphism is explained in detail in this answer to another question.
The proof that they are isomorphic works by just calculating the dimensions on both sides and finding that they are the same. The dimensions are infinite cardinals. That proof is given here.