Suficient condition for tensor product of vector spaces.. Can anyone help me showing the following:
Let $E$, $F$, $G$ and $H$ vector spaces and $\varphi:E\times F\rightarrow G$ a bilinear map. If for every $\psi:E\times F\rightarrow H$ bilinear there is an unique linear map $f:G\rightarrow H$ such that $\psi=f\circ \varphi$ then $(G, \varphi)$ is a tensor product of $E$ and $F$.
The definition I have for tensor product of vector spaces is:
Definition: Let $E, F$ and $G$ be vector spaces and $\varphi:E\times F\rightarrow G$ a bilinear map. We say the pair $(G, \varphi)$ is a tensor product of $E$ and $F$ if:
(i) $\textrm{im}(\varphi)=G$.
(ii) For every $\psi:E\times F\rightarrow H$ bilinear, where $H$ is an arbitrary vector space, there is linear map $f:G\rightarrow H$ such that $\psi=f\circ \varphi$.
So using this definition it suffices showing the map $\varphi:E\times F\rightarrow G$ is surjective for solving my problem...
Any help will be welcome..Thanks
 A: (i) is not correct, it should be $\langle \mathrm{im}(\phi) \rangle = G$.
In order to solve the "exercise" (which in fact, is the equivalence between your wrong definition (which should be seen as a characterization) and the correct definition), you only have to show (i). To do that, apply the given universal property to see that the inclusion $\langle \mathrm{im}(\phi) \rangle \hookrightarrow G$ induces a bijection $\hom(G,H) \to \hom(\langle \mathrm{im}(\phi) \rangle,H)$ for every $H$. This implies that $\langle \mathrm{im}(\phi) \rangle \to G$ is an isomorphism (Yoneda Lemma; any argument (especially the ones which will appear in the other answers) are repetitions of the proof of the Yoneda Lemma).
A: Your confusion stems from a lack of understanding of the unconventional notation used by your book (Greub's Multilinear Algebra, 2nd ed). 
In p. 2, the book states: "We shall denote $\text{Im}\,\varphi$ the subspace of $G$ generated by $S$", where $S$ is the usual image of $\varphi$ and $G$ is the codomain of $\varphi$.
The author does indeed formulate the way you've mentioned. However, the first thing in the book (p. 1) is to show that the image of a bilinear map is not in general a subspace. The definition you cite is in p. 4; so just read the previous pages, which are worthwhile.
