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My notes ask me to confirm the result:

$\mathrm{Hom}(V \otimes W, U) \cong \mathrm{Hom}(V, \mathrm{Hom}(U,W))$ as $\mathbb C$-representations of a finite group $G$.

But it seems to be incorrect. $\chi_{\mathrm{Hom}(V \otimes W, U)} = \overline{\chi_{V \otimes W}} \chi_U = \overline{\chi_V}\ \overline{\chi _ W} \chi_U$, whilst $\chi_{\mathrm{Hom}(V, \mathrm{Hom}(U,W))} = \overline{\chi_V} \chi_{\mathrm{Hom}(U,W)} = \overline{\chi_V} \ \overline{\chi_U} \chi_W$. So it looks like $U$ and $W$ should be switched in the RHS of the isomorphism.

Am I right?

Thanks

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    $\begingroup$ Yes, you are right. $\endgroup$ – Qiaochu Yuan Mar 22 '12 at 20:17
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Yes, you're right. This is a version of hom-tensor adjunction. Instead of equality of characters you could also write down an ismorphism of vector spaces: in one direction it should send $f: V\otimes W \to U$ to the map that sends $v\in V$ to $w \mapsto f(v\otimes w) \in \hom(W,U)$.

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An alternative proof (for all fields is) ${\rm Hom}(X\otimes Y,Z)\cong (X\otimes Y)^*\otimes Z\cong (X^* \otimes Y^*)\otimes Z\cong X^* \otimes (Y^*\otimes Z)\cong {\rm Hom}(X,{\rm Hom}(Y,Z)).$

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