Brauer images retain the surviving primitive idempotents

Let $\varphi :A\twoheadrightarrow B$ and $y^2=y$ in $B$. The cases $y=0,1$ lift by $0,1$. Otherwise choose $x$ above $y$. Linear dependence among $1,x,\dots ,x^{\dim A}$ gives $m(x)=0$ for a nonzero polynomial $m$. Since $m(y)=m(0)(1-y)+m(1)y=0$ and both idempotents are nonzero, multiplication by them forces $m(0)=m(1)=0$. Write $m(T)=g(T)(T-1{)}^s$ with $s\ge 1$ maximal, so $g(1)\mathrm{\ne }0$ and $g(0)=0$. Bézout supplies $ug+v(T-1{)}^s=1$. Set $q=ug$; then $q(q-1)=-uvm$, $q(0)=0$ and $q(1)=1$. Therefore $q(x{)}^2=q(x)$ and $\varphi (q(x))=q(y)=y$. Polynomial evaluation is valid even in noncommutative $A$ because powers of one element commute.

For an idempotent $e\in A$, the corner map $eAe\to \varphi (e)B\varphi (e)$ is onto: a target $\varphi (e)b\varphi (e)$ has preimage $eae$ if $\varphi (a)=b$. Apply step 1.1 using identity $e$ in this corner. If $e$ is primitive and $\varphi (e)\mathrm{\ne }0$ were not, a nontrivial idempotent in its target corner would lift to $u\in eAe$ with $u\mathrm{\ne }0,e$. Then $e=u+(e-u)$ would be an orthogonal nontrivial decomposition, impossible. Conversely a primitive target idempotent lifts by step 1.1, then decomposition by [F3] has exactly one nonzero image, which supplies a primitive lift.

Homomorphisms preserve sums and orthogonality. Apply step 2.1 to the surjection $A_R\to B_R$ in [F2] and delete zero images in a decomposition of one. For the second clause decompose $i$ in $A_Q$, which contains $A_P$. A summand $j\le i$ is primitive in $i{A}_{Q}i$ exactly when primitive in $A_Q$, since $j(i{A}_{Q}i)j=j{A}_{Q}j$. The same image argument gives the primitive decomposition of ${\mathrm{Br}}_Q(i)$; if all images vanish it is the empty decomposition. No primitivity of ${\mathrm{Br}}_Q(i)$ itself is asserted when $i$ is only primitive in $A_P$.