Decompositions of the identity into primitive central idempotents are unique

If $k=0$ or $\mathrm{\ell }=0$ then $1=0$ in $A$, so $A$ has exactly one element by [F1], namely $0$; then no nonzero idempotent exists and the two families are empty, $k=\mathrm{\ell }=0$, and the empty map is the required bijection. Assume henceforth that $k,\mathrm{\ell }\ge 1$.

For central idempotents $e,f\in A$ the products $ef$ and $e(1-f)$ are central idempotents with $ef+e(1-f)=e$ and $ef\cdot e(1-f)=0$: centrality gives $(ef)a=e(fa)=e(af)=(ea)f=(ae)f=a(ef)$ and $(1-f)a=a-af=a-fa=a(1-f)$ for all $a\in A$; idempotence gives $(ef{)}^2=efef=e^2{f}^2=ef$ and $(1-f{)}^2=1-2f+f^2=1-f$, hence $e(1-f)e(1-f)=e^2(1-f{)}^2=e(1-f)$; distributivity gives $ef+e(1-f)=e(f+1-f)=e$ and $ef\cdot e(1-f)=e^{2}f(1-f)=0$.

If $e$ is primitive and $f$ is a central idempotent, then $ef\in \{0,e\}$: by step 1.2, $e=ef+e(1-f)$ exhibits $e$ as a sum of the two orthogonal central idempotents $ef$ and $e(1-f)$, so primitivity forces $ef=0$ or $e(1-f)=0$; in the second case step 1.2 gives $ef=ef+0=ef+e(1-f)=e$.

Fix $j\in \{1,\dots ,\mathrm{\ell }\}$. Since $1=e_1+\cdots +e_k$, distributivity and centrality give $f_j=f_j\cdot 1={\sum }_i{f}_j{e}_i={\sum }_i{e}_i{f}_j$, and step 2.1 gives $e_i{f}_j\in \{0,e_i\}$ for every $i$; since $f_j\mathrm{\ne }0$, the set $S_j:=\{i:e_i{f}_j=e_i\}$ is nonempty and $f_j={\sum }_{i\in S_j}{e}_i$.

For every $j$ the set $S_j$ is a singleton: its elements $e_i$ are pairwise orthogonal nonzero central idempotents summing to $f_j$ by step 3.1, so if $S_j$ contained distinct $i\mathrm{\ne }{i}^{\mathrm{\prime }}$ then $f_j=e_i+{\sum }_{i^{\mathrm{\prime }\mathrm{\prime }}\in S_j,i^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\ne }i}{e}_{i^{\mathrm{\prime }\mathrm{\prime }}}$ would exhibit $f_j$ as a sum of the two orthogonal central idempotents $e_i$ and $f_j-e_i$, both nonzero since multiplying the second sum by the nonzero idempotent $e_{i^{\mathrm{\prime }}}$ retrieves $e_{i^{\mathrm{\prime }}}$, contradicting primitivity of $f_j$; hence $S_j=\{i(j)\}$ and $f_j=e_{i(j)}$.

The map $j\mapsto i(j)$ is injective: if $i(j)=i(j^{\mathrm{\prime }})=i$ with $j\mathrm{\ne }{j}^{\mathrm{\prime }}$, then $e_i=e_i^2=f_j{f}_{j^{\mathrm{\prime }}}=0$ by idempotence, centrality and pairwise orthogonality of the $f$'s, contradicting $e_i\mathrm{\ne }0$.

The map $j\mapsto i(j)$ is surjective: from $f_j=e_{i(j)}$ for all $j$ and both decompositions, ${\sum }_j{e}_{i(j)}=1={\sum }_i{e}_i$; multiplying this equality by the central idempotent $e_i$ and using pairwise orthogonality of the $e$'s gives $e_i$ on the left when $i$ lies in the image of $j\mapsto i(j)$ and $0$ otherwise, while the right side is $e_i\mathrm{\ne }0$; hence every $i$ lies in the image.

Steps 5.1 and 5.2 make $j\mapsto i(j)$ a bijection $\{1,\dots ,\mathrm{\ell }\}\to \{1,\dots ,k\}$ with $f_j=e_{i(j)}$ for all $j$ by step 4.1; its inverse $\sigma$ satisfies $e_i=f_{\sigma (i)}$ for all $i$, and $k=\mathrm{\ell }$. Since the families were arbitrary, the decomposition of $1$ into primitive central idempotents is unique up to order, as claimed.