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Let $G$ be a finite group.

In their paper Some Remarks On the Structure of Mackey functors , Greenlees and May define a functor:

$R: GMod \rightarrow M[G]$ where $GMod$ is the category of finite left $G$-modules and $M[G]$ is the category of $G$ Mackey functors by:

$RV(G/H) = V^H$ where $V^H= \{v\in V| h(v) = v \hspace{.2cm} \forall h\in H\}$ and $V$ is a $G$ module.

Question: In Theorem 12, why are $coker(\eta)$ and $ker(\eta)$ in $\mathcal{A}$?

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I'm fairly sure this question would be ok for MO. You may get better access to experts in this stuff over there. – Justin Young Sep 16 '12 at 6:59

You should think of the lexicographic ordering on pairs $(j,k)$ as giving an ordering on Mackey functors (called type) as well as on the subgroups $H \leq G$ (which I'll also call type, even though they don't). In this language, a Mackey functor has type $(j,k)$ exactly when it vanishes on all subgroups of type less than $(j,k)$, but it doesn't vanish on subgroups of type $(j,k)$.

Now, suppose that $type(M)=(j,k)$. We have a canonical Mackey functor $R(M(G/H_{j,k}))$ of type $(j,k)$ [see footnote] and with the same value on $G/H_{j,k}$, and this receives a map $\eta$ from $M$ which is the identity on $G/H_{j,k}$. Neither the source nor the target of $\eta$ ever evaluated nontrivially on subgroups of type less than $(j,k)$, and because of the above bolded fact, the kernel and cokernel of $\eta$ also evaluate trivially on $G/H_{j,k}$. Thus, they have type strictly greater than $(j,k)$, so they are in $\mathscr{A}$ by assumption.

footnote: This is where we use the assumption that if $type(H)=(j,k)$, then $type(J)_1<j$ for all $J \lneq H$.

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