Green correspondence identity boundaries

Suppose $H=G$. Restriction leaves every action unchanged. The map $kG{\otimes }_{kG}U\to U$, $g\otimes u\mapsto gu$, is balanced, since $(ga)u=g(au)$. Its inverse is $u\mapsto 1\otimes u$: the tensor relation $g\otimes u=1\otimes gu$ verifies the other composite. Thus induction is naturally isomorphic to the identity. Also $G\setminus H$ is empty, so $\mathcal{X}=\mathcal{Y}=\mathrm{\varnothing }$ and $\mathcal{Z}=\{Q:Q\le P\}$.

For a nonzero indecomposable input in this domain, the identity module has just itself as an indecomposable summand: any proposed nonzero complement contradicts indecomposability. Thus both distinguished assignments are identity maps and the errors vanish. This also follows from the empty-family clauses of F1. AC is retained for the general theorem used to identify these assignments, not for the explicit tensor maps in 1.1.

If $P=1$, then every element normalizes $P$, so $N_G(P)=G$ and necessarily $H=G$. A nonzero indecomposable has vertex $1$ exactly when it is relatively $1$-projective: one implication is part of vertex minimality, and the other follows because $1$ has no proper subgroup. By F3 these are exactly the projective indecomposables. F2 therefore restricts the identity map from 2.1 to precisely that domain. Zero remains outside the domain, and zero errors cause no exception. This proves both boundary assertions.