Tensor-Hom Adjunction

Transclude of 2-Modules#^60fa98

Lemma

If $U,V$ and $W$ are $kG$-modules, then

\mathrm{Hom}_{kG}(U\otimes V,W)\cong\mathrm{Hom}_{kG}(U,V^*\otimes W). $$ ^rtw9uw

\begin{proof} Note that there is an isomorphism $U^{\ast }\otimes V\cong {\mathrm{Hom}}_k(U,V)$, which maps $\phi \otimes v$ to ${\varphi }_v$ where ${\phi }_v:u\mapsto \phi (u)v$. So there is

$$\begin{array}{rl}{\mathrm{Hom}}_k(U\otimes V,W)& \cong (U\otimes V{)}^{\ast }\otimes W\\ & \cong U^{\ast }\otimes (V^{\ast }\otimes W)\\ & \cong {\mathrm{Hom}}_k(U,V^{\ast }\otimes W).\end{array}$$

If $U$ and $V$ are $kG$-modules, then ${\mathrm{Hom}}_k(U,V)$ is also a $kG$-module by $(g\phi )(u):=g(\phi (g^{-1}u))$. For any $\phi \in {\mathrm{Hom}}_k(U,V)$, if $\phi$ is an element of ${\mathrm{Hom}}_{kG}(U,V)$, then we have that $(g\phi )(gu)=g(\phi (u))=\phi (g(u))=\phi (gu)$ for all $u\in U$ and $g\in G$ (see here). It yields that $g\phi =\phi$ and so ${\mathrm{Hom}}_{kG}(U,V)$ consists of the elements of ${\mathrm{Hom}}_k(U,V)$ which are fixed by $G$.

Since an isomorphism will map the fixed points of $G$ in the one space to the fixed points in the other space, we obtain that ${\mathrm{Hom}}_{kG}(U\otimes V,W)\cong {\mathrm{Hom}}_{kG}(U,V^{\ast }\otimes W)$. \end{proof}