A global block of defect D determines a local block of defect D

Take a maximal $B$-pair $(P,a)$ from F2. By F3 its subgroup $P$ is a defect group. Apply F4 to $P$ and $D$ in both directions: their orders agree, and $P$ is conjugate to $D$. Conjugate the pair to have subgroup literally $D$. Then F2 gives $e={\mathrm{Br}}_D(f)$ as the sum of one $N$-orbit of primitive central idempotents of $k{C}_G(D)$. It is nonzero, idempotent and $N$-invariant, hence central in $kN$.

Suppose $u$ is a central idempotent of $kN$ beneath $e$. Since $D⊴N$, F1 gives $u={\mathrm{Br}}_D(u)\in k{C}_G(D)$. It is central in that algebra, since $C_G(D)\le N$. Thus $u$ is a sum of a subset of the primitive idempotents in step 1.1. Centrality in $kN$ makes this subset $N$-stable, and one transitive orbit has only the empty and whole stable subsets. Hence $u=0$ or $u=e$. So $e$ is primitive in $Z(kN)$ and defines a block $b$. Moreover ${\mathrm{Br}}_D(e)=e\mathrm{\ne }0$. If $S>D$ were a $p$-subgroup of $N$ with ${\mathrm{Br}}_S(e)\mathrm{\ne }0$, F7 and $C_G(S)\subseteq C_G(D)$ would give ${\mathrm{Br}}_S(f)={\mathrm{Br}}_S(e)\mathrm{\ne }0$, contradicting F4 for $f$. Therefore F4, now in $N$, proves $D$ is a defect group of $b$.

Since $C_G(D)\le N$, F5 defines $b^G$ and gives ${\lambda }_{b^G}(f)={\lambda }_b(e)=1$. F6 therefore identifies $b^G=B$. Conversely, if a block $c$ of $kN$ with defect group $D$ induces to $B$, F5 gives $1={\lambda }_B(f)={\lambda }_c(e)$. Since $e$ is primitive by step 2.1, F6 forces $c=b$. This proves uniqueness in the stated defect-$D$ domain as well as the promised existence. For $D=1$, $N=G$ and the construction gives $e=f$; the same holds whenever $N=G$ by F1. All chosen pairs, subgroups and idempotent subsets lie in finite sets; no AC or stronger restriction-summand theorem was used.