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\title{Miscellaneous notes}

\begin{document}

\maketitle

For a finite $G$-set $K$, let $\ul{K} = \SI^{\infty}_G(K_+)$. 
Think of $G$-spectra as tensored and cotensored over unbased
$G$-spaces via $E\odot X = E\sma X_+$ and $F(X,E) = F(X_+,E)$. 

\begin{lem} For finite $G$-sets $K$ and $L$, $F_G(\ul K, \ul L)$
is equivalent to $S_G\odot (K\times L)$.
\end{lem}
\begin{proof} Since $\ul K$ is self-dual, we have
\[ F_G(\ul K,\ul L)\simeq F_G(\ul K, S_G) \sma \ul L\simeq \ul K\sma \ul L
\simeq S_G\odot (K\times L) \qedhere \]
\end{proof}

This determines the zeroth space up to equivalence as $Q_G(K\times L)_+$ 
and identifies its nonequivariant group of components as $\bZ[K\times L]$. 
This is the nonequivariant components with $G$ acting.  Is the following
pair of maps helpful?
\[\xymatrix@1{
 F_G(\ul K,\ul L) & \SI^{\infty}_GF_G(\ul K,\ul L)_0 \ar[l] \ar[r] &
 \SI^{\infty}_G \bZ[K\times L] }\]
 
\[ F_G(\ul L,\ul M)\sma F_G(\ul K,\ul L)\rtarr F_G(\ul K,\ul M) \]

is

\[ S_G\odot (L\times M \times K\times L) \rtarr S_G\odot (K\times M) \]

but there is no reason to think that this is $S_G\odot (-)$ as a $G$-map. 
 

\end{document} 