I've already seen a construction which ensures the existence of tensor product to a family of vector spaces $V_1,\cdots, V_k$. But i've been asked to show a diferent construction. Here it is:

Let $V = \mbox{Hom}^k(V_1,\cdots,V_k, \mathbb F)$ and $\varphi: V_1\times \cdots \times V_k \rightarrow V^*$ given by $\varphi(v_1,\cdots,v_k)(f)=f(v_1,\cdots,v_k)$ for every $f\in V$ and for every $v_j\in V_j, j=1,\cdots, k$. Prove that the pair $(\varphi, \mbox{Im}(\varphi))$ is a tensor product to $V_1,\cdots, V_k.$

Note: Here $\mbox{Hom}^k(V_1,\cdots,V_k,\mathbb F)$ is the set of all multilinear transformations: $F: V_1\times\cdots\times V_k \rightarrow \mathbb F.$ Also, each $V_j$ is not supposed to be finite dimensional.

The book that i'm using defines tensor product via universal property. So i must show that $(\varphi, \mbox{Im}(\varphi))$ satisfies the universal property of tensor products. But i'm struggling to show that.

Give, $W$ any vector space, and any $\phi\in \mbox{Hom}(V_1,\cdots,V_k,W)$, i've seen that this is equivalent to show that $\mbox{Im}\varphi$ generates $V^*$ and to show that there exists $f\in \mbox{Hom}(V^*,W)$ such that $\phi = f \circ \varphi$. The existence of a such $f$ can be done as a user answered above. But i still cannot show that $\mbox{Im}\varphi$ generates $V^*$

Any help would be much appreciated.

  • $\begingroup$ There seems to be a typo: "...for every $f \in V^*$..." should presumably be "...for every $f \in V$..." $\endgroup$ – Stephen Nov 10 '17 at 13:21
  • $\begingroup$ @Stephen you're correct. I'll edit $\endgroup$ – math.h Nov 10 '17 at 13:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.