# Hom and Tensor Product of Linear Maps

I am reading Claudio Procesi's book on Lie groups and on page 105 there is something I don't understand.

Let $U,V,W$ be vector spaces. Let us consider the product space $\hom(V,W) \times \hom(U,V)$ together with the map

$$f : \hom(V,W) \times \hom(U,V) \rightarrow \hom(U,W)$$

that sends a pair $(v, \varphi)$ to $v \circ \varphi$. Suppose I have an elementary tensor $v \otimes \varphi$ in $\hom(V,W) \otimes\hom(U,V)$. By the universal property of tensor products there is a unique linear map

$$L : \hom(V,W) \otimes \hom(U,V) \rightarrow \hom(U,W)$$

such that on elementary tensors, $L(v\otimes \varphi) = v \circ \varphi$.

Now 1.5.1 of pg 105 claims that for any $u \in U$, we have that

$$(v \otimes \varphi)(u) = v \langle \varphi | u\rangle$$

where $\langle \varphi | u\rangle$ is defined on page 16 to be the value of the linear form $\varphi$ on $u$. I am confused as to how they obtained that equality above - is it really an equality? I don't see it directly but I think it comes from the fact that

$$\hom(U,V) \cong V \otimes U^\ast.$$

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I agree with Brad - on the left hand side of your penultimate equation, you have $v \otimes \phi$, an element of the tensor product of two hom spaces. It doesn't make sense to "evaluate" this at $u \in U$, so you should imagine a silent $L$ or regard the whole equation as a definition. – m_t_ Oct 9 '12 at 14:46
@mt_ What if we now take $U = W = V$, would this make sense? – user38268 Oct 9 '12 at 14:49
no matter what U,W,V are, literally speaking $v\otimes \phi$ isn't a function that can be applied to objects of type $u \in U$. The only natural way I can see to make sense of it is to regard the equation as a definition. – m_t_ Oct 9 '12 at 14:54
@mt_ Thanks, I agree that the equality does not make sense. That's why I was confused in the first place. – user38268 Oct 10 '12 at 8:49

I think, it wants to simply mean $$\left[L(v \otimes \varphi)\right](u) = v(\varphi ( u))$$ well.. this notation $\langle\varphi|u\rangle$ now confused me too, as $\varphi$ is not a linear form. The $L$ should be understood, because without it the LHS gives syntax error, as mt_ wrote in the comment..