The trivial-defect-group and identity-normalizer boundaries

If $N_G(D)=G$, restriction from $G\times G$ to itself changes no module. A block bimodule $B$ occurs in itself. It cannot occur in another block $B^{\mathrm{\prime }}$: the central idempotent of $B$ acts as identity on $B$ and as zero on $B^{\mathrm{\prime }}$, hence on all its summands. Thus F3 gives $B^G=B$, and F1's map and inverse are the identity on the set of blocks with defect group $D$. This proves both directions of the correspondence directly.

For $D=1$, every group element normalizes $D$, so step 1.1 applies. The numerical defect is ${\log }_p\mathrm{\mid }D\mathrm{\mid }=0$, and F2 says exactly these blocks are full matrix algebras with one projective simple module. If there are no such blocks, the identity has empty domain and codomain: its injectivity condition and the assertion that every codomain element has a preimage are both vacuous. The same reasoning handles an empty defect-$D$ set whenever $D$ is normal. The zero algebra is not inserted as a block. All these identity and idempotent calculations are choice-free.