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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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    @mt_ Thanks, I a$g$ree that the equality does not make sense. That's why I was confused in the first place.2012-10-10

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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..