A separability idempotent constructs the balanced relative tensor product

Since ${\sigma }_B$ is a bimodule map and $m_B{\sigma }_B=1_B$, the element $e_B={\sigma }_B{\eta }_B$ satisfies $m_B{e}_B={\eta }_B$ and the centrality equation $((m_B\otimes 1_B)\circ (1_B\otimes e_B))=((1_B\otimes m_B)\circ (e_B\otimes 1_B)):B\to B\otimes B\mathrm{.}$ Associators are inserted in the displayed composite according to source and target. This equation says, in element notation, ${\sum }_{i}b{x}_i\otimes y_i={\sum }_i{x}_i\otimes y_{i}b$ for $e_B={\sum }_i{x}_i\otimes y_i$.

The centrality equation gives $p_{X,Y}(r_X\otimes 1_Y)=p_{X,Y}(1_X\otimes l_Y)$: on an elementary tensor, both maps apply the equality ${\sum }_{i}b{x}_i\otimes y_i={\sum }_i{x}_i\otimes y_{i}b$ between the middle factors. Thus $p_{X,Y}$ is balanced. If $f:X\otimes Y\to Z$ is balanced, then $f{p}_{X,Y}(x\otimes y)={\sum }_{i}f(x{x}_i\otimes y_{i}y)={\sum }_{i}f(x\otimes x_i{y}_{i}y)=f(x\otimes y),$ where the first equality is the definition of $p$, the second is balancing, and the third uses $m_B{e}_B={\eta }_B$, the unit action on $Y$. This calculation is an equality of composites of action maps, so it does not depend on element notation. Taking $f=p_{X,Y}$ proves $p_{X,Y}^2=p_{X,Y}$.

Write $X\otimes Y$ as a finite direct sum of pairwise nonisomorphic simple objects using [F2]. A nonzero morphism between simple objects has kernel either zero or the whole source and image either zero or the whole target by [F4]; nonzero excludes the latter kernel and former image, so the morphism is an isomorphism. Thus distinct simple summands have zero Hom spaces, and $\mathrm{End}(S)$ is a finite-dimensional division algebra over $k$. For any $u\in \mathrm{End}(S)$, its minimal polynomial splits over the algebraically closed field $k$; the equation ${\prod }_j(u-{\lambda }_{j}1)=0$ and absence of zero divisors imply $u={\lambda }_{j}1$ for some root ${\lambda }_j$, so $\mathrm{End}(S)=k$. Therefore $p_{X,Y}$ is a finite block matrix of $k$-linear idempotents on the multiplicity spaces. For each such idempotent $P:V\to V$, every $v\in V$ is $Pv+(v-Pv)$ with $Pv\in \mathrm{im}P$ and $v-Pv\in \ker P$, while $\mathrm{im}P\cap \ker P=0$; hence $V=\mathrm{im}P\oplus \ker P$ and $P$ splits. Taking these splittings over all blocks gives $X\otimes Y\stackrel{q_B}{\to }P\stackrel{i}{\to }X\otimes Y$ with $i{q}_B=p_{X,Y}$ and $q_{B}i=1_P$. Since $p_{X,Y}$ equalizes the two action maps, $q_B$ does too: apply the monomorphism $i$ to the two composites from $X\otimes B\otimes Y$ and use step 2.1. Hence $q_B$ is balanced.

If $f:X\otimes Y\to Z$ is any balanced map, step 2.1 gives $f{p}_{X,Y}=f$. Therefore $\tilde{f}=fi:P\to Z$ satisfies $\tilde{f}{q}_B=fi{q}_B=f{p}_{X,Y}=f$. It is unique because $q_{B}i=1_P$: any $g:P\to Z$ with $g{q}_B=f$ must satisfy $g=g{q}_{B}i=fi$. Thus $P$ has the balanced universal property.

Any other splitting of the same idempotent has the same universal property. The unique maps between the two representing objects that preserve their universal balanced maps are inverse, since each composite preserves that same universal map and uniqueness in step 4.1 forces it to be the identity. This proves canonical independence of the splitting.