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

  • $\begingroup$ Is the above true for any vector space $\,H\,$ ? $\endgroup$
    – DonAntonio
    Jun 29, 2013 at 15:39
  • $\begingroup$ Which definition of the tensor product do you use? $\endgroup$ Jun 29, 2013 at 15:40
  • $\begingroup$ I was writing exactly that @BorisNovikov =D $\endgroup$
    – PtF
    Jun 29, 2013 at 15:41
  • $\begingroup$ That's right @DonAntonio $\endgroup$
    – PtF
    Jun 29, 2013 at 15:42

2 Answers 2


(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).

  • $\begingroup$ Thanks for the tip.. As to $(i)$ note that $\textrm{span}\{\varphi(x, y):x\in E, y\in F\}=G$ if and only if $\textrm{im}(\varphi)=G$... $\endgroup$
    – PtF
    Jun 29, 2013 at 16:02
  • $\begingroup$ @PtF: I don't understand your comment. Your claim is false. $\endgroup$ Jun 29, 2013 at 16:29
  • $\begingroup$ Well: If $g\in G$ then $g\in \textrm{span}\{\varphi(x, y):x\in E, y\in F\}$, that is, $\displaystyle g=\sum_{\alpha\in I}\alpha \varphi(x, y)$ for some scalars $\alpha$. Since $\varphi(x, y)\in \textrm{im}(\varphi)$ and $\textrm{im}(\varphi)$ is closed for vector operations we have $g\in \textrm{im}(\varphi)$ hence $G\subset \textrm{im}(\varphi)$ and the equality $\textrm{im}(\varphi)=G$ holds. $\endgroup$
    – PtF
    Jun 29, 2013 at 17:51
  • $\begingroup$ Conversely if $g\in \textrm{span}\{\varphi(x, y):x\in E, y\in F\}$ then $\displaystyle g=\sum_{\alpha\in I}\alpha_I\varphi(x, y)$ and $g\in \textrm{im}(\varphi)=G$, that is, $\textrm{span}\{\varphi(x, y):x\in E, y\in F\}\subset G$. Now if $g\in G=\textrm{im}(\varphi)$ then $g=\varphi(x, y)$ for some $x\in E$ and $y\in F$ hence $g\in \textrm{span}\{\varphi(x, y):x\in E, y\in F\}$. Therefore, $\textrm{span}\{\varphi(x, y):x\in E, y\in F\}=G$. $\endgroup$
    – PtF
    Jun 29, 2013 at 17:52
  • $\begingroup$ $\phi$ is not linear, it is bilinear. It's image is not a subspace (unless $\dim(E),\dim(F) \leq 1$). $\endgroup$ Jun 29, 2013 at 20:02

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.


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