Module functor adjoints and separable-monad splittings in Mod(C)

Choose representatives $a$ of the finitely many simple objects of $\mathcal{M}$. Put $R(n)={⨁}_a{\mathrm{Hom}}_{\mathcal{N}}(F(a),n){\otimes }_{k}a$ and $L(n)={⨁}_a{\mathrm{Hom}}_{\mathcal{N}}(n,F(a){)}^{\ast }{\otimes }_{k}a$. Since $\mathrm{End}(a)=k$, decomposition into these simples gives natural bijections ${\mathrm{Hom}}_{\mathcal{M}}(m,Rn)={\mathrm{Hom}}_{\mathcal{N}}(Fm,n)$ and ${\mathrm{Hom}}_{\mathcal{M}}(Ln,m)={\mathrm{Hom}}_{\mathcal{N}}(n,Fm)$. Define the units and counits to represent the identity maps under these bijections. The two triangle identities follow by applying either bijection: in each case the composite corresponds to the original identity. Thus these are actual ordinary adjunctions, with no exactness or representability assumption left implicit.

Let $(T,\mu ,\eta ,\sigma )$ be a separable monad on $\mathcal{M}$. Define ${\mathcal{M}}^T$ to consist of pairs $(x,a)$ with $a:Tx\to x$, $a{\eta }_x=1$ and $aT(a)=a{\mu }_x$; its arrows $h:(x,a)\to (y,b)$ obey $ha=bT(h)$. The underlying additive functor $T$ is exact because $\mathcal{M}$ is semisimple. Thus the kernel and cokernel of $h$ carry induced $T$-actions, by exactness and the universal properties, satisfying the two algebra equations. They give the categorical kernel and cokernel in ${\mathcal{M}}^T$. The canonical comparison from coimage to image is the underlying isomorphism and preserves actions; its inverse preserves them as well. Therefore ${\mathcal{M}}^T$ is abelian and the forgetful functor $U$ is exact and faithful. Its free left adjoint is $L_T(x)=(Tx,{\mu }_x)$, with unit ${\eta }_x$ and counit at $(x,a)$ equal to $a$. The two triangles are respectively ${\mu }_{x}T{\eta }_x=1$ and $a{\eta }_x=1$.

Fix a dual pair $X,X^{\vee }$ with evaluation $X\otimes X^{\vee }\to \mathbf{1}$ and coevaluation $\mathbf{1}\to X^{\vee }\otimes X$. Define $r_{n,X}:R(n\odot X)\to Rn\odot X$ by the following natural bijections in $m$: ${\mathrm{Hom}}_{\mathcal{M}}(m,R(n\odot X))\cong {\mathrm{Hom}}_{\mathcal{N}}(Fm,n\odot X)\cong {\mathrm{Hom}}_{\mathcal{N}}(F(m\odot X^{\vee }),n)\cong {\mathrm{Hom}}_{\mathcal{M}}(m\odot X^{\vee },Rn)\cong {\mathrm{Hom}}_{\mathcal{M}}(m,Rn\odot X)$. The middle identification uses $f_{m,X^{\vee }}$ and the other identifications use the adjunction or the displayed duality. Every arrow is invertible, so Yoneda gives an invertible module comparison. To verify its pentagon, apply $\mathrm{Hom}(m,-)$ to the two paths for $X,Y$. Both paths move $Y^{\vee }$ and then $X^{\vee }$ across Hom, apply $f$ to the resulting action, then move them back. The equality of these composites is exactly the module pentagon for $f$, since evaluation of the dual of $X\otimes Y$ is the consecutive evaluation of $Y$ and $X$ with the associators. For $X=\mathbf{1}$ the same comparison reduces to the module unit equation for $f$. This proves all module-functor axioms for $R$.

For $(x,a)$ put $s_{(x,a)}=T(a){\sigma }_x{\eta }_x:x\to Tx$. Its composite with $a$ is $aT(a){\sigma }_x{\eta }_x=a{\mu }_x{\sigma }_x{\eta }_x=a{\eta }_x=1$. It is an algebra map to $(Tx,{\mu }_x)$: indeed $sa=T(a){\sigma }_x{\eta }_{x}a=T(a){\sigma }_{x}T(a){\eta }_{Tx}=T(a)T^2(a){\sigma }_{Tx}{\eta }_{Tx}=T(a)T({\mu }_x){\sigma }_{Tx}{\eta }_{Tx}=T(a){\sigma }_x{\mu }_x{\eta }_{Tx}=T(a){\sigma }_x$, while ${\mu }_{x}T(s)={\mu }_x{T}^2(a)T({\sigma }_x)T({\eta }_x)=T(a){\mu }_{Tx}T({\sigma }_x)T({\eta }_x)=T(a){\sigma }_x{\mu }_{x}T({\eta }_x)=T(a){\sigma }_x$. Here the middle equalities use, in order, naturality, the algebra equation, the two bimodule equations for $\sigma$, and the unit equations. Naturality of $s$ in algebra maps follows by the same naturality of $\sigma ,\eta$ and $ha=bT(h)$. Thus every algebra is a natural retract of its free algebra.

Under the bijections of step 2.1, the image of the adjunction counit ${ϵ}_n:FRn\to n$ is its tensor translate, so the equality $({ϵ}_n\odot 1_X)f_{Rn,X}F(r_{n,X})={ϵ}_{n\odot X}$ holds. This is precisely module-naturality of $ϵ$. For the unit, apply the adjunction bijection to its two proposed module-naturality composites; both correspond to the identity of $F(m\odot X)$ after $f$ and its inverse cancel. Faithfulness of the bijection proves their equality. Hence both triangles from step 1.1 are identities of module-natural transformations. For $L$, use the dual pair with evaluation $X^{\mathrm{♭}}\otimes X\to \mathbf{1}$ and coevaluation $\mathbf{1}\to X\otimes X^{\mathrm{♭}}$ in the identical argument with the Hom adjunction reversed: $\mathrm{Hom}(L(n\odot X),m)\cong \mathrm{Hom}(n,Fm\odot X^{\mathrm{♭}})\cong \mathrm{Hom}(Ln,m\odot X^{\mathrm{♭}})\cong \mathrm{Hom}(Ln\odot X,m)$. Its middle comparison is that of $F$. Evaluation of tensor products gives the module pentagon, the unit pair gives its unit equation, and applying this bijection proves module-naturality of its unit and counit.

For each $x$, $L_T(x)$ is projective, since ${\mathrm{Hom}}_{{\mathcal{M}}^T}(L_{T}x,-)={\mathrm{Hom}}_{\mathcal{M}}(x,U-)$ is exact. Hence all objects are projective by step 2.2. Let $g$ be the sum of the finitely many simples of $\mathcal{M}$. Every object of ${\mathcal{M}}^T$ is a quotient of a finite sum of copies of $L_{T}g$: first its underlying object is a summand of a finite sum of copies of $g$, and then its action is an epimorphism from its free object. Also $\mathrm{End}(L_{T}g)$ is finite dimensional, being a subspace of an underlying Hom space in $\mathcal{M}$. Every object has finite length, since a strict chain of subobjects remains a strict chain after the faithful exact functor $U$ and $U(x,a)=x$ has finite length. With all objects projective, all short exact sequences split, so each object is a finite sum of simples. Each simple is a summand of $L_{T}g$ because its quotient from a finite sum of $L_{T}g$ splits; the finite length of $L_{T}g$ bounds the number of such simples. Thus ${\mathcal{M}}^T$ is finite semisimple with finite dimensional Hom spaces.

Write $t_{x,X}:T(x\odot X)\to Tx\odot X$ for the module comparison. Define $(x,a)\odot X=(x\odot X,(a\odot 1)t_{x,X})$. The fact that $\mu$ and $\eta$ are module-natural proves its two algebra equations. The associativity and unit constraints inherited from $\mathcal{M}$ are algebra maps by the pentagon and unit equation for $t$, and obey their own pentagon and triangle because $U$ is faithful. Thus ${\mathcal{M}}^T$ is a finite semisimple right module category. The displayed action makes $U$ a module functor, and $t$ gives $L_T$ its module structure. Module-naturality of $\eta ,\mu$ verifies the adjunction unit and counit. Module-naturality of $\sigma ,\eta$ and of the action $a$ verifies that $s$ from step 2.2 is module-natural as well. Consequently $L_T⊣U$ is a separable adjunction, induces exactly $T$ with its multiplication and unit, and supplies the required splitting.