Right module internal ends and Eilenberg–Watts equivalence

Write $I$ for representatives of the simple objects of $\mathcal{C}$. The object $[m,n]={⨁}_{i\in I}{\mathrm{Hom}}_{\mathcal{M}}(m\odot i,n){\otimes }_{k}i$ represents the stated functor: the decomposition $X={⨁}_i\mathrm{Hom}(i,X)\otimes i$ identifies both Hom spaces, naturally in $X$. Evaluation $v_n:m\odot [m,n]\to n$ is the corresponding universal morphism. The adjunction $T⊣R$, $T(X)=m\odot X$, $R(n)=[m,n]$, is thus explicit, and $R$ is exact since each coefficient $\mathrm{Hom}(m\odot i,-)$ is exact in a semisimple category. It is faithful by the generator hypothesis: if $h:n\to n^{\mathrm{\prime }}$ is nonzero, precomposing with an epimorphism $TX\to n$ gives a nonzero map, and adjunction identifies its image under $R$ as nonzero.

Let $F:{\mathrm{Mod}}^l(A)\to {\mathrm{Mod}}^l(B)$ be right exact and a right module functor, and put $Q=F(A)$. It has its given left $B$-action. Apply $F$ to the left $A$-linear map $\mu :A\otimes A\to A$ and use $F(A\otimes A)\cong Q\otimes A$ to obtain $r_Q:Q\otimes A\to Q$. The algebra identities for $\mu$ and coherence of the module comparison give the right action identities for $r_Q$; since $F(\mu )$ is a left $B$-map, the two actions commute. For a left $A$-module $P$, define $Q{\otimes }_{A}P$ as the cokernel of $r_Q\otimes 1-1\otimes l_P$ on $Q\otimes A\otimes P$; the commuting left $B$-action descends. Tensoring and cokernels are right exact, so this is a right exact right module functor.

For any $X$, choose its dual $X^{\vee }$ with evaluation $X\otimes X^{\vee }\to \mathbf{1}$ and coevaluation $\mathbf{1}\to X^{\vee }\otimes X$. Rigidity gives natural bijections $\mathrm{Hom}(Y,R(n\odot X))=\mathrm{Hom}(m\odot Y,n\odot X)=\mathrm{Hom}(m\odot (Y\otimes X^{\vee }),n)=\mathrm{Hom}(Y,R(n)\otimes X)$. Yoneda gives $R(n\odot X)\cong R(n)\otimes X$. Applying this construction successively to $X,Y$ and applying it once to $X\otimes Y$ give the same bijection: the evaluation of a tensor product is the composite of the two evaluations in reverse order, and their units and counits satisfy the two zigzag identities. This proves the module pentagon for this comparison; evaluation against $\mathbf{1}$ proves its triangle. Evaluation and the adjunction unit are module-natural by the same representing bijections.

The two maps $A\otimes A\otimes U(P)\rightrightarrows A\otimes U(P)$, namely $\mu \otimes 1$ and $1\otimes l_P$, have coequalizer $l_P:A\otimes U(P)\to P$ in left $A$-modules. Indeed, if $h$ equalizes them, its unique factor is $h(\eta \otimes 1)$; the equalizing identity says both that this is left $A$-linear and that its composite with $l_P$ equals $h$. Since $F$ is right exact, applying it preserves this coequalizer. Using $F(A\otimes X)\cong Q\otimes X$, the resulting parallel maps are exactly $r_Q\otimes 1$ and $1\otimes l_P$. Thus their universal maps give a natural isomorphism $Q{\otimes }_{A}P\cong F(P)$. Its module compatibility follows on the free source from the module comparison of $F$, and follows on $P$ by composing with the epimorphism from that free source.

Set $A=R(m)$. Its unit $\eta :\mathbf{1}\to A$ represents $1_m$ and its multiplication $\mu :A\otimes A\to A$ represents $m\odot (A\otimes A)\cong (m\odot A)\odot A\stackrel{v_m\odot 1}{\to }m\odot A\stackrel{v_m}{\to }m$. Associativity follows because both triple composites represent evaluation three times, with equality given by the module pentagon; the two unit identities represent evaluation with an identity inserted. In particular on $\mathrm{Hom}(\mathbf{1},A)=\mathrm{End}(m)$ this multiplication is $a\cdot b=b\circ a$. The action $A\otimes R(n)\to R(n)$ represents $m\odot (A\otimes R(n))\to m\odot R(n)\to n$, first using $v_m$, then $v_n$. The same triple-evaluation computation proves that it is a left action. Equivalently it is the action of the monad $RT=A\otimes -$ obtained from the adjunction, so its associativity and unit also follow by applying the two adjunction triangles. The resulting comparison $K:n\mapsto R(n)$ is a right module functor to left $A$-modules; $K(TX)=A\otimes X$ with its free left action.

A module-natural transformation $F\to F^{\mathrm{\prime }}$ evaluated at $A$ is a $B$-$A$ bimodule map: $B$-linearity is automatic and $A$-linearity follows by naturality at $\mu$ and module-naturality. Conversely a bimodule map $Q\to Q^{\mathrm{\prime }}$ induces maps of the two coequalizer diagrams of step 2.2 and hence a module-natural transformation. Evaluation at $A$ recovers it, using $Q{\otimes }_{A}A\cong Q$ by the action. Every transformation is determined by its component on $A$: the module constraint determines it on $A\otimes X$, and naturality along the free epimorphism determines it on $P$. These operations are inverse and preserve identities and composition. This proves the full Eilenberg–Watts equivalence, with source algebra on the right of $Q$ and target algebra on its left.

Kernels and cokernels of maps of $A$-modules are their kernels and cokernels in $\mathcal{C}$, with induced actions: tensoring by $A$ is exact, hence preserves the defining kernel and cokernel diagrams. Thus the forgetful functor is exact and faithful. Every left module $P$ has a free presentation $A\otimes Y\to A\otimes X\to P\to 0$: take the action $A\otimes U(P)\to P$, whose underlying map is split by $\eta \otimes 1$; its kernel is another module and is likewise the quotient of a free module. For a free source, the map on Hom spaces induced by $K$ is the bijection ${\mathrm{Hom}}_A(A\otimes X,Rn)={\mathrm{Hom}}_{\mathcal{C}}(X,Rn)={\mathrm{Hom}}_{\mathcal{M}}(TX,n)$.

Every object of $\mathcal{M}$ is a quotient of $T(X)$ by the module-generator hypothesis (for the sum of simples, take $X$ to be a finite sum of units); applying the same argument to its kernel gives a presentation $T(Y)\to T(X)\to n\to 0$. Exactness of $R$ sends it to a free presentation of $K(n)$. Applying $\mathrm{Hom}(-,n^{\mathrm{\prime }})$ and ${\mathrm{Hom}}_A(-,K(n^{\mathrm{\prime }}))$ identifies the two kernels of the maps out of the free terms by step 4.1. Hence $K$ is fully faithful. Conversely, lift the first map of a free presentation of $P$ uniquely to $T(Y)\to T(X)$ by step 4.1 and take its cokernel $n$ in $\mathcal{M}$. Exactness gives $K(n)\cong P$. Thus $K$ is the asserted right module equivalence. The zero category is included by taking $m=0$ and the zero algebra, whose unit identity holds on the zero object.

To obtain the left-action version, view a left $\mathcal{C}$-module as a right ${\mathcal{C}}^{\mathrm{r}\mathrm{e}\mathrm{v}}$-module by $n◃X=X\odot n$. The module pentagon is the same pentagon with the tensor inputs reversed, so all preceding constructions apply. An algebra in ${\mathcal{C}}^{\mathrm{r}\mathrm{e}\mathrm{v}}$ gives an algebra $A$ in $\mathcal{C}$ with the same underlying multiplication morphism $A\otimes A\to A$, reading tensor positions in the order of $\mathcal{C}$; a left module over it in ${\mathcal{C}}^{\mathrm{r}\mathrm{e}\mathrm{v}}$ is exactly a right $A$-module in $\mathcal{C}$. Thus the fully faithful and essentially surjective comparison of step 5.1 becomes the stated equivalence to right modules, and the projection comparison of step 2.1 becomes $[m,X\odot n]\cong X\otimes [m,n]$. In this convention the internal-end multiplication represents $A\odot (A\odot m)\to A\odot m\to m$, and on ordinary endomorphisms is usual composition. In particular $[m,X\odot m]\cong X\otimes A$ as right $A$-modules, with the free right action. No separability assertion entered any of these arguments.