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Let $V,W$ be linear spaces and $\varphi: V\times W \rightarrow \mbox{Hom}(V^*,W)$ given by: $\varphi(v,w)(f) = f(v)w$ for every $v\in V, w\in W, f\in V^*$. Supposing that $\dim(V) <\infty$, prove that the pair $(\varphi, \mbox{Hom}(V^*,W))$ is a tensor product to $V,W$.

Notation: $\mbox{Hom}(V,W)$ is the set of all linear maps from $V$ to $W$.

Well, i've shown that $\varphi(v,w)$ is indeed an element of $\mbox{Hom}(V^*,W)$ for every $v\in V,w\in W$, by proving that $\varphi(v,w)$ is indeed linear. Also, i've shown that $\varphi$ is a bilinear map.

Therefore, by universal property we guarantee a unique $\overline \varphi\in \mbox{Hom}(V\otimes W, \mbox{Hom}(V^*,W))$ such that $\varphi(v,w) = \overline \varphi (v\otimes w) $ for every $v\in V , w\in W$.

Since any other tensor product to $V,W$ is given by $(f\circ \otimes, f(V\otimes W))$, where $f$ is bijection, if we prove that $\overline \varphi$ is actually bijective, we are done. We proceed to show this:

$\overline \varphi$ is injective: given $u\in \mbox {Ker}(\overline \varphi)$, we choose an expression $u = \sum_{i=1}^m v_i\otimes w_i$ where $v_i$ and $w_i$ are linearly independent (minimal rank). We argue that $m=0$. If not, since $v_1 \neq 0 $, we choose $f_1\in V^*$ such that $f_1(v_j) = \delta_{1,j}$. Hence, $$ 0 = \overline \varphi(u) = \sum_{j=1}^{m}\overline \varphi (v_j\otimes w_j) = \sum_{j=1}^{m}\varphi(v_j,w_j).$$ Evaluating at $f_1$, then:

$$0 = \overline\varphi(u)(f_1) = \sum_{j=1}^{m}\varphi(v_j,w_j)(f_1) = \sum_{j=1}^{m}f_1(v_j)w_j = f_1(v_1)w_1$$ and since $w_1$ is l.i., it must be $f_1(v_1) = 0$, a contradiction with our choice of $f_1\in V^*$. Therefore, $m=0$ and hence $u = 0$.

Question: How to prove surjection? Since $\mbox{dim}(V)<\infty$, we have $\mbox{dim}(V) = \mbox{dim}(V^*)$. Will this ensure that $\mbox{dim}(V\otimes W) = \mbox{dim}(\mbox{Hom}(V^*,W))$? How?

Thanks in advance!

user2345678
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    Note: if you already know that i) $\mathcal{L}(V,W) \cong V^\otimes W$ and ii) when $\dim(V)< \infty$, $V\cong V^$, then you know that

    $\mathcal{L}(V^,W)\cong \mathcal{L}(V,W) \cong V^\otimes W\cong V\otimes W.$

    In addition, the composition of these isomorphisms should equal $\varphi$.

     – AnonymousCoward Nov 09 '17 at 17:41
  • @AnonymousCoward

    Thank you for your answer. Would you mind expliciting a function which shows the isomorphism between $\mathscr L(V,W)$ and $V^*\otimes W?$

     – user2345678 Nov 09 '17 at 18:37
  • One way to describe the isomorphism is to fix a basis of $V$ and a dual basis $e_i^{\vee}$ for $V^*$. Then for $A\in \mathcal{L}(V,W)$, $A\mapsto \sum_i e_i^{\vee}\otimes A(e_i)$. If you expand each $A(e_i)$ in terms of a basis of $W$, then this amounts to writing matrix entries for $A$. A coordinate free map in the other direction, similar to the one you describe above is given in this post: https://math.stackexchange.com/a/679600/565 – AnonymousCoward Nov 09 '17 at 19:07

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