Module-functor categories are finite semisimple

The ordinary linear functor category $\mathcal{E}={\mathrm{Fun}}_k(\mathcal{M},\mathcal{N})$ is finite semisimple: evaluation on representatives $m_a$ of the simples of $\mathcal{M}$, followed by decomposition in simples $n_b$ of $\mathcal{N}$, identifies it with the category of finite matrices of vector spaces $V_{ba}=\mathrm{Hom}(n_b,F(m_a))$. A natural transformation is exactly a matrix of linear maps; conversely the matrix defines the functor by additive extension. The category $\mathcal{D}$ acts on $\mathcal{E}$ by $(X⊠Y)\cdot F=R_Y^{\mathcal{N}}F{R}_X^{\mathcal{M}}$. Indeed $R_Y{R}_W=R_{W\otimes Y}$ and $R_Z{R}_X=R_{X\otimes Z}$, giving the product $(X\otimes Z)⊠(W\otimes Y)$ in $\mathcal{D}$. The action coherence follows by applying the two module pentagons; its two unit equations are the two module triangles. Thus $K\cdot -$ is an exact linear monad on $\mathcal{E}$ with multiplication induced by that of $K$.

A left $K$-action on $F$ consists of components ${\theta }_{i,m}:F(m\odot i)\odot i^{\vee }\to F(m)$. By duality these correspond bijectively to $c_{m,i}:F(m\odot i)\to F(m)\odot i$; the inverse correspondence is ${\theta }_{i,m}=(1\odot {\mathrm{ev}}_i)(c_{m,i}\odot 1)$. Finite semisimplicity extends these maps uniquely and naturally from simple $i$ to every $X$: decompose $X$ into simples, use additivity on the source, and the dual decomposition on the dual factor. Independence of bases follows from the identity ${\sum }_a{g}_a{f}_a=1_X$ for dual inclusion/projection maps. Expanding $\theta (\mu \cdot 1)=\theta (1\cdot \theta )$ on the $(i,j)$ summand and using the formula for $\mu$ in F1 gives, after taking mates, $c_{m,i\otimes j}=(c_{m,i}\odot 1_j)c_{m\odot i,j}$ with the module associators inserted: the sum over $f,g$ inserts exactly $\sum gf=1_{i\otimes j}$. The unit action equation gives $c_{m,\mathbf{1}}=1$ under the unitors. Conversely those two equations give the action equation on every summand of $K\otimes K$, hence on the full direct sum.

These initially lax comparisons are invertible by rigidity. Choose a dual $X^{\mathrm{♭}}$ with $b:\mathbf{1}\to X\otimes X^{\mathrm{♭}}$ and $d:X^{\mathrm{♭}}\otimes X\to \mathbf{1}$. An inverse to $c_{m,X}$ is the composite $F(m)\odot X\stackrel{F(1_m\odot b)\odot 1_X}{\to }F(m\odot X\odot X^{\mathrm{♭}})\odot X\stackrel{c_{m\odot X,X^{\mathrm{♭}}}\odot 1_X}{\to }F(m\odot X)\odot X^{\mathrm{♭}}\odot X\stackrel{1\odot d}{\to }F(m\odot X)$. For its composite after $c_{m,X}$, naturality with $b$, followed by the product equation of step 2.1, leaves the zigzag $(1_X\otimes d)(b\otimes 1_X)=1_X$. For the composite before $c_{m,X}$ use naturality with $d$ and the same product equation; the other zigzag and the unit equation leave $1_{F(m\odot X)}$. A map of $K$-actions is exactly a natural transformation commuting with all ${\theta }_i$, equivalently all $c_X$. Thus the category of $K$-modules in $\mathcal{E}$ is precisely ${\mathrm{Fun}}_{\mathcal{C}}(\mathcal{M},\mathcal{N})$, including its morphisms.

Kernels and cokernels of $K$-module maps exist and are their underlying kernels and cokernels in $\mathcal{E}$, since $K\cdot -$ is exact. They carry the induced actions by the universal properties. For modules $(P,{\rho }_P),(Q,{\rho }_Q)$ and an arbitrary underlying arrow $t:P\to Q$, put $\mathrm{Av}(t)={\rho }_Q(1_K\cdot t)(1_K\cdot {\rho }_P)(e\cdot 1_P)$. In symbolic notation for this composite it is $\mathrm{Av}(t)(p)={\sum }_a{x}_{a}t(y_{a}p)$, where $e={\sum }_a{x}_a\otimes y_a$ denotes the morphism $e$, not actual elements. Centrality says $\sum b{x}_a\otimes y_a=\sum x_a\otimes y_{a}b$, so associativity of the actions gives $\mathrm{Av}(t)(bp)=b\mathrm{Av}(t)(p)$ as an equality of the corresponding composites. Thus it is a module map. If $t$ already is a module map, the identity $\mu e=\eta$ gives $\mathrm{Av}(t)=t$. If $q:Q\to P$ is a module epimorphism and $qt=1_P$, the same formula gives $q\mathrm{Av}(t)=\mathrm{Av}(qt)=1_P$. Every underlying epi splits in $\mathcal{E}$, so all module epis split.

Hom spaces of modules are finite dimensional subspaces of those in $\mathcal{E}$. Every module has finite length, since its subobject chains give strict subobject chains in its underlying object of $\mathcal{E}$. Therefore the splitting in step 4.1 implies that every module is a finite direct sum of simples. To prove finiteness of the simple set, let $g$ be the sum of the finitely many simples of $\mathcal{E}$. The free module $K\cdot g$ has finite length. Every module is a quotient of a finite sum of copies of $K\cdot g$: its underlying object is a summand of a finite sum of copies of $g$, apply the free functor and then the action epimorphism. Each simple is therefore a summand of a finite sum of $K\cdot g$ and hence is isomorphic to a simple constituent of $K\cdot g$. There are only finitely many. Steps 3.1–5.1 prove the finite semisimplicity assertion. Under F2 this also proves it for the correctly oriented bimodule categories, without assuming separability of their representing algebras in advance.

A right module endofunctor $F$ of the regular category satisfies $F(X)=F(\mathbf{1}\otimes X)\cong F(\mathbf{1})\otimes X$, naturally and compatibly with its comparison. Conversely $a\mapsto (a\otimes -)$ is a right module functor, with the associator as comparison. The same comparison shows that any module-natural transformation is $h\otimes 1$ for its component $h$ at $\mathbf{1}$, and every $h$ defines one. Thus evaluation at $\mathbf{1}$ and $a\mapsto (a\otimes -)$ are inverse equivalences. The identity functor corresponds to $\mathbf{1}$, whose endomorphism algebra is $k$. Hence the identity 1-morphism of the regular module is simple, which is the asserted simple-unit condition.