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?