TITLE: Alternative definition for tensor product which nowhere occurs (to my knowledge)
QUESTION [6 upvotes]: Every vector space $V$ could be embedded into $V^{\ast}$ (see here) after choosing a basis, for a given vector $v \in V$ denote this embedding by $v^{\ast}\in V^{\ast}$. Now for given vector spaces $V_1, \ldots, V_k$ over some field $F$, let $V = \{ \varphi : V_1 \times \ldots \times V_k \to F \mbox{ multilinear } \}$. Why not define the tensor product 
of $V_1, \ldots, V_k$ simply as $T = \{ \varphi^{\ast} \mid \varphi\in V\}$. Then the universal property is obviously fulfilled, for if we
define $\pi : V_1 \times \ldots \times V_k \to T$ by $\pi(v_1, \ldots, v_k) = \Phi \in V^{\ast}$ with
$$
 \Phi(\varphi) = \varphi(v_1, \ldots, v_k).
$$
Then if we have some multilinear $\varphi : V_1 \times \ldots \times V_k \to F$ define the linear map $h_{\varphi} : T \to F$ by
$$
 h_{\varphi}(\Phi) = \Phi(\varphi)
$$
and we have
$$
 h_{\varphi}(\pi(v_1, \ldots, v_k)) = \varphi(v_1, \ldots, v_k)
$$
i.e. it factors through $T$ by $\pi$ and $h_{\varphi}$. Then everything works out quite easily, no nasty "quotient constructions", it even appears too simple for me...
I have nowhere seen this definition? So why not define it that way? Have I overlooked something? Note that we do not rely on reflexivity here, as $T$ does not has to be all of $V^{\ast}$, but just those elements that arise from elements of $V$ (the image of the embedding). Maybe the universal property breaks down because the linear map is not unique, but I do not see other choices for it?

REPLY [4 votes]: Let me mention something which has not been explicitly stated in any of the other answers: your construction is emphatically wrong for infinite-dimensional vector spaces (not just, you have to tweak it or do more work prove it works, but it just doesn't give the right answer at all).
This is easiest to see by considering the case $k=1$.  The tensor product of just a single vector space $V_1$ should, of course, be (isomorphic to) $V_1$ itself, since a multilinear map on the one-fold product $V_1$ is the same thing as a linear map on $V_1$.  However, your definition of $T$ would make it isomorphic to $V$, which is just the dual space $V_1^*$ in this case.  If $V_1$ is infinite-dimensional, then $V_1^*$ is not isomorphic to $V_1$!
(The basic thing that goes wrong with your argument in this case is that you haven't checked uniqueness of your $h_\varphi$.  You have only constrained the value of $h_\varphi$ on elements $\Phi$ of the form $\pi(v_1)$, but not every element of $T$ has this form, since $V=V_1^*$ is bigger than $V_1$.)