Draft lesson.

Separable monads and the Karoubi completion of free modules

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

A free module is often easier to describe than a general module. Separability provides a natural retraction from a free module onto every module. We prove the abstract monad version, identify the corresponding completion of the free-module category, and retain the maps on morphisms and natural transformations. The examples distinguish a natural splitting from the mere existence of enough free objects.

1. Monad modules and the Kleisli category

Fix universes for the categories, Hom sets and choices used below. Let $T:\mathcal C\to\mathcal C$ have natural unit $\eta:1\to T$ and multiplication $\mu:T^2\to T$, satisfying $\mu_XT\eta_X=1_{TX}=\mu_X\eta_{TX}, \qquad \mu_XT\mu_X=\mu_X\mu_{TX}. \tag{1.1}$ A $T$-module is $(X,a)$, where $a:TX\to X$, $a\eta_X=1$ and $aTa=a\mu_X$. Its maps satisfy $fa=bTf$. Write $\mathcal C^T$ for this category, $U$ for its forgetful functor, and $F(X)=(TX,\mu_X)$.

Tensor actions and absolute algebra presentations, Section 2 proves $F\dashv U$. The inverse Hom rules are $g\mapsto g\eta_X$ and $f\mapsto aTf$. Its counit at $(X,a)$ is $a:FUX\to X$, a module map.

Define $\operatorname{Kl}(T)$ to have the objects of $\mathcal C$, and maps $f:X\to Y$ given by underlying maps $f:X\to TY$. Its identity is $\eta_X$; write its composition as $g\star f=\mu_ZTg\,f\quad (f:X\to TY,\ g:Y\to TZ). \tag{1.2}$ The two units follow from (1.1) and naturality of $\eta$. For associativity, both $(h\star g)\star f$ and $h\star(g\star f)$ reduce to $\mu_WT\mu_WT^2h\,Tg\,f$. For the latter first use naturality $Th\,\mu_Z=\mu_{TW}T^2h$, then $\mu_W\mu_{TW}=\mu_WT\mu_W$. This verifies the category laws explicitly.

Proposition 1.1. The functor $J:\operatorname{Kl}(T)\longrightarrow\mathcal C^T, \quad JX=FX,\quad Jf=\mu_YTf \tag{1.3}$ is fully faithful, with image the free modules.

Proof. The displayed map is a module map by the associativity and naturality of $\mu$, equivalently by the free-module Hom rule. Its inverse on Hom sets sends $g:FX\to FY$ to $g\eta_X$. One composite is $\mu_YTf\,\eta_X=\mu_Y\eta_{TY}f=f$. The other is $\mu_YTg\,T\eta_X=g\mu_XT\eta_X=g$, using the module-map equation for $g$.

It takes the Kleisli identity to $\mu_XT\eta_X=1_{TX}$. Moreover $\mu_ZT(\mu_ZTg\,f) =\mu_ZT\mu_ZT^2g\,Tf =\mu_ZTg\,\mu_YTf,$ where the final equality uses monad associativity and naturality for $g:Y\to TZ$. Thus it preserves composition. $\square$

2. Separable multiplication gives a natural free retraction

The monad is separable if there is a natural $\sigma:T\to T^2$ with $\mu_X\sigma_X=1,\qquad T\mu_X\,\sigma_{TX} =\sigma_X\mu_X =\mu_{TX}T\sigma_X. \tag{2.1}$ The last two equations are the two module-linearity equations for its multiplication section. No monoidal structure on $\mathcal C$ is required.

Theorem 2.1. For every module $(X,a)$, the map $\xi_{(X,a)}=Ta\,\sigma_X\eta_X:X\longrightarrow TX \tag{2.2}$ is a module map to $FUX$, natural in module maps, and $a\xi_{(X,a)}=1_X$.

Proof. The splitting equation is $a\xi=aTa\,\sigma_X\eta_X =a\mu_X\sigma_X\eta_X=a\eta_X=1.$ For module linearity compare $\xi a$ and $\mu_XT\xi$. Naturality of $\eta,\sigma$, the module equation for $a$, and then (2.1) give $\begin{aligned} \xi a &=Ta\,\sigma_XTa\,\eta_{TX}\\ &=Ta\,T^2a\,\sigma_{TX}\eta_{TX}\\ &=Ta\,T\mu_X\,\sigma_{TX}\eta_{TX}\\ &=Ta\,\sigma_X\mu_X\eta_{TX} =Ta\,\sigma_X. \end{aligned} \tag{2.3}$ The other route gives $\begin{aligned} \mu_XT\xi &=\mu_XT^2a\,T\sigma_XT\eta_X\\ &=Ta\,\mu_{TX}T\sigma_XT\eta_X\\ &=Ta\,\sigma_X\mu_XT\eta_X =Ta\,\sigma_X. \end{aligned} \tag{2.4}$ Here the second line is naturality of $\mu$ for $a:TX\to X$. Thus $\xi a=\mu_XT\xi$.

For a module map $f:(X,a)\to(Y,b)$, substitute $fa=bTf$ into $\xi_Yf=Tb\,\sigma_Y\eta_Yf =Tb\,T^2f\,\sigma_X\eta_X =Tf\,Ta\,\sigma_X\eta_X =Tf\,\xi_X.$ This proves naturality and completes the theorem. $\square$

Consequently every module is a retract of its specified free module $FUX$, with inclusion $\xi_X$ and quotient $a$. The associated free-module projector is $\xi_Xa:TX\to TX$.

For a tensor monad $T=A\otimes-$, Separable algebra objects and absolute balanced tensor, Sections 2 and 5 constructs exactly these sections from the multiplication section of $A$.

3. The converse uses natural Hom retractions

A functor $G:\mathcal D\to\mathcal C$ is separable if its maps on Hom sets have retractions $H_{A,B}$ which are natural under morphisms in $\mathcal D$. Precisely, $H(Gf)=f,\qquad H(Gb\,h\,Ga)=bH(h)a. \tag{3.1}$ Here $h:GA\to GB$ can be any underlying morphism, while $a,b$ are morphisms of $\mathcal D$. This does not require $H$ to preserve composition of arbitrary underlying morphisms.

Lemma 3.1. For an adjunction $F\dashv G$, $G$ is separable if and only if its counit has a natural section in $\mathcal D$.

Proof. Given a section $\xi_A:A\to FGA$ of $\varepsilon_A$, put $H(h)=\varepsilon_BFh\,\xi_A. \tag{3.2}$ Naturality of $\xi,\varepsilon$ gives (3.1); for a morphism $f$ in $\mathcal D$, it gives $H(Gf)=f\varepsilon_A\xi_A=f$.

Conversely define $\xi_A=H_{A,FGA}(\eta_{GA}). \tag{3.3}$ By naturality in the second Hom variable and the adjunction triangle, $\varepsilon_A\xi_A=H(G\varepsilon_A\,\eta_{GA})=H(1_{GA})=1_A$. For $f:A\to B$ in $\mathcal D$, naturality of $H$ in both variables and naturality of $\eta$ give $FGf\,\xi_A =H(GFGf\,\eta_{GA}) =H(\eta_{GB}Gf) =\xi_Bf.$ Thus the section is natural. The unit, counit and their triangles are those proved in Passing maps across an adjunction, Section 2. $\square$

Theorem 3.2. A monad is separable if and only if $U:\mathcal C^T\to\mathcal C$ is separable. Multiplication sections and natural sections of the free-module counit correspond bijectively.

Proof. Theorem 2.1 and Lemma 3.1 prove one direction. Conversely a natural section $\xi$ of the counit defines $\sigma_X=U(\xi_{FX}):TX\to T^2X$. It is natural because $\xi$ is natural on the free maps $Ff$. Its splitting equation is $\mu_X\sigma_X=1$.

The fact that $\xi_{FX}$ is a module map gives $\sigma_X\mu_X=\mu_{TX}T\sigma_X$. Naturality of $\xi$ on the module counit $\mu_X:F(TX)\to FX$ gives the other equation $\sigma_X\mu_X=T\mu_X\,\sigma_{TX}$. These are all of (2.1).

To verify the inverse correspondences, start with $\sigma$ and evaluate (2.2) at $FX=(TX,\mu_X)$. Its value is $T\mu_X\,\sigma_{TX}\eta_{TX}=\sigma_X\mu_X\eta_{TX}=\sigma_X$. In the other direction naturality on $a:FUX\to(X,a)$ says $\xi_Xa=Ta\,\sigma_X$. Precompose with $\eta_X$ and use $a\eta_X=1$ to recover exactly (2.2). $\square$

4. Completing a category by retracts

For any category $\mathcal D$, define $\operatorname{Kar}(\mathcal D)$ to have objects $(X,p)$ with $p^2=p$. A morphism to $(Y,q)$ is an arrow $f:X\to Y$ with $qf=f=fp$. Composition is the original composition, and the identity at $(X,p)$ is $p$. These equations check closure under composition and the identity laws. The embedding $I(X)=(X,1_X)$ is fully faithful.

The arrows both represented by $p$ exhibit $(X,p)$ as a retract of $IX$. An idempotent $e$ on $(X,p)$ satisfies $pe=e=ep$; it splits through $(X,e)$, using $e$ in both directions. Hence $\operatorname{Kar}(\mathcal D)$ is idempotent complete. The same corner descriptions occur in Retracts and stabilization of formal objects, Sections 1 and 3; here we also prove the functorial universal property.

Theorem 4.1. For idempotent-complete $\mathcal E$, restriction along $I$ is an equivalence of functor categories $\operatorname{Fun}(\operatorname{Kar}(\mathcal D),\mathcal E) \simeq\operatorname{Fun}(\mathcal D,\mathcal E). \tag{4.1}$ This includes all natural transformations.

Proof. Given $H:\mathcal D\to\mathcal E$, choose splittings $HX\xrightarrow{q_p}E_p\xrightarrow{i_p}HX$ of $H(p)$. For $p=1_X$ choose the identity splitting. Define $\overline H(X,p)=E_p,\qquad \overline H(f)=q_qH(f)i_p. \tag{4.2}$ The identity is $q_pH(p)i_p=q_pi_p=1$. For composable corner maps, inserting $i_qq_q=H(q)$ does not change their composite, since $gq=g$ and $qf=f$. This proves functoriality and gives an extension of $H$.

For a natural transformation $\theta:H\to K$, its extension has component $\overline\theta_{(X,p)} =q_p^K\theta_Xi_p^H. \tag{4.3}$ Naturality of $\theta$ for $p$ says $K(p)\theta_X=\theta_XH(p)$. Substitution in (4.2) verifies the naturality square for every corner map $f$. The same splitting equations prove preservation of identities and vertical composition of transformations.

Conversely any functor on the completion sends the two formal retract arrows to a splitting of its value on $p$. A transformation between two such functors is determined by its components on $IX$: naturality for the inclusion and projection forces (4.3), using those image splittings. That formula also constructs the unique extension of every transformation on the restriction. Thus restriction is fully faithful. Construction (4.2) proves essential surjectivity, giving (4.1). Different splitting choices identify by the unique splitting isomorphisms, naturally in all corner maps; their three-choice comparisons compose. $\square$

5. Splitting a module projector

Proposition 5.1. If $\mathcal C$ is idempotent complete, so is $\mathcal C^T$, for any monad $T$.

Proof. Let $p:(X,a)\to(X,a)$ be an idempotent module map, and split its underlying map as $p=iq,qi=1_Y$. Define $b=q\,a\,Ti:TY\longrightarrow Y. \tag{5.1}$ Since $pa=aTp$, we have $ib=paTi=aTi,\qquad bTq=qaTp=qa. \tag{5.2}$ The equality $qaTp=qa$ follows by composing $pa=aTp$ with $q$ and using $qp=q$.

Naturality of $\eta$ gives $ib\eta_Y=a\eta_Xi=i$; cancel the split monomorphism $i$ to get $b\eta_Y=1$. Also $ibTb=aTiTb=aT(ib)=aTa\,T^2i =a\mu_XT^2i=aTi\,\mu_Y=ib\mu_Y.$ Cancel $i$ again. Thus $b$ is a module action. Equations (5.2) say exactly that $i,q$ are module maps. They split $p$ in $\mathcal C^T$. No preservation of kernels, images or chosen splittings by $T$ was assumed. $\square$

6. All modules are the completed free category

Theorem 6.1. For a separable monad on any $\mathcal C$, the fully faithful free-module functor induces an equivalence $\operatorname{Kar}(\operatorname{Kl}(T)) \simeq\operatorname{Kar}(\mathcal C^T). \tag{6.1}$ If $\mathcal C$ is idempotent complete, it gives $\operatorname{Kar}(\operatorname{Kl}(T))\simeq\mathcal C^T$.

Proof. A fully faithful functor remains fully faithful on Karoubi completions: its Hom bijection restricts to exactly the corner equations for the two projectors. Every module is a retract of a free one by Theorem 2.1. An object of $\operatorname{Kar}(\mathcal C^T)$ is in turn a formal retract of a module. Compose the two retract inclusions and the two projections. Their quotient-inclusion composite is the identity, so it is a retract of a free object. Its free-object idempotent is transported to a unique Kleisli idempotent by full faithfulness of $J$. Splitting that formal idempotent gives the required object in the source of (6.1), proving essential surjectivity. Proposition 5.1 and Theorem 4.1 identify the second completion with $\mathcal C^T$ when $\mathcal C$ is idempotent complete. $\square$

Here are actual maps for the comparison. The arrow $\xi_X:X\to TX$ is a Kleisli idempotent: $\xi_X\star\xi_X=\mu_XT\xi_X\,\xi_X =\xi_Xa\xi_X=\xi_X. \tag{6.2}$ It corresponds under $J$ to the free-module projector $\xi_Xa$. Define $K(X,a)=(X,\xi_X),\qquad K(f)=\xi_Yf:X\longrightarrow TY. \tag{6.3}$ Its target and source corner equations follow from $\mu_YT\xi_Y=\xi_Yb$, $fa=bTf$, and $a\xi_X=b\xi_Y=1$. Explicitly $\xi_Y\star K(f)=K(f)$ and $K(f)\star\xi_X=\xi_YbTf\,\xi_X=\xi_Yfa\xi_X=K(f)$. The identity is $K(1_X)=\xi_X$, the identity of its Karoubi object. For $f:X\to Y,g:Y\to Z$, $K(g)\star K(f) =\xi_ZcTg\,\xi_Yf =\xi_Zgb\xi_Yf =\xi_Zgf=K(gf).$ Thus $K$ is a functor. The maps $a,\xi_X$ identify its completed free image with $(X,a)$, naturally in module maps. These formulas also describe the inverse comparison after formal retract completion; they retain the chosen natural section rather than forgetting the corner projectors.

7. Choices and the limits of the criterion

Two separability sections give two counit sections $\xi,\xi'$ and hence two functors $K,K'$. At a module $(X,a)$, the underlying Kleisli arrow $\xi'_X$ is an isomorphism $(X,\xi_X)\longrightarrow(X,\xi'_X), \tag{7.1}$ with inverse $\xi_X$. Indeed $\xi_X\star\xi'_X=\xi_Xa\xi'_X=\xi_X$ and $\xi'_X\star\xi_X=\xi'_X$, exactly the two corner identities. It satisfies the corner equations by the same calculation. Naturality on $f$ is the equation $\xi'_Y\star(\xi_Yf)=(\xi'_Yf)\star\xi_X=\xi'_Yf$. For a third section the two successive comparisons compose to its own $\xi''_X$, so the comparisons are coherent.

The converse to Theorem 6.1 is not asserted. Having enough free retracts does not produce a natural section of the specified counit. In Theorem 3.2 that natural section is essential.

For clarity, let $R$ be a commutative ring and $A$ a unital central $R$-algebra. On $\operatorname{Mod}(R)$ the tensor monad $A\otimes_R-$ is separable if and only if $A$ is separable. A multiplication section of $A$ supplies the monad section by tensoring with $1_M$. Conversely, given a natural monad section, its value on $R$ is an $R$-linear map $s:A\to A\otimes_R A$. For each $m\in M$, naturality on $R\to M,1\mapsto m$ gives $\sigma_M(a\otimes m)=s(a)\otimes m$. Pure tensors generate, so this determines the section. Its splitting equation on $R$ is $ms=1$, and its two equations (2.1), evaluated on $R$, are precisely the left and right $A$-linearity of $s$. This proves the equivalence with its actual maps. It does not require base flatness.

8. Four graded exercises with complete solutions

Exercise 8.1 — An idempotent rational-localization monad (intermediate)

On abelian groups take $T(M)=\mathbb Q\otimes_{\mathbb Z}M$, with its usual unit and multiplication. Prove its multiplication invertible, identify its modules and explain why every module is already free.

Solution. Multiplication $\mathbb Q\otimes\mathbb Q\to\mathbb Q$ has inverse $q\mapsto q\otimes1$. To check this on $q\otimes s/n$, multiplication by $n$ on the tensor is invertible through its first rational factor. Both $q\otimes s/n$ and $qs/n\otimes1$ become $qs\otimes1$ after multiplying by $n$, hence are equal. Pure tensors generate, proving the inverse equation. Tensoring with $M$ proves $\mu_M$ invertible.

More generally, if $\mu$ is invertible, both monad unit maps $T\eta_X,\eta_{TX}$ equal $\mu_X^{-1}$. For a module action $a$, the equation $Ta\,T\eta_X=1$ then gives $Ta=\mu_X$. Naturality of $\eta$ gives $\eta_Xa=Ta\,\eta_{TX}=\mu_X\mu_X^{-1}=1$. Thus $a$ and $\eta_X$ are inverse. The action is an isomorphism from $FUX$, so every module is free. For separability put $\sigma=\mu^{-1}$. Cancelling the invertible $\mu_X$ from associativity gives $T\mu_X=\mu_{TX}$. Hence $T\mu_X\,\sigma_{TX}=1_{T^2X} =\sigma_X\mu_X =\mu_{TX}T\sigma_X,$ which proves both linearity equations for this section.

For this particular monad its modules are exactly uniquely divisible abelian groups, or rational vector spaces. An action gives multiplication by $1/n$, inverse to multiplication by $n$, using its unit and associativity; hence division is unique. Conversely define rational scalar multiplication on a uniquely divisible group by the unique solution of $ny=sx$. Common denominators verify addition and composition of scalars and give the action on $\mathbb Q\otimes M$. Maps are exactly rational-linear maps. The free comparison is consequently already an equivalence, before adding retracts.

Exercise 8.2 — A separable monad can have a nonfree module (intermediate)

On $\operatorname{Vect}_k$, let $T=(k\times k)\otimes_k-$. Construct a module which is a retract of a free module but is not free, and exhibit its Kleisli idempotent.

Solution. Put $A=k\times k$, with orthogonal units $e_1,e_2$. The tensor $e_1\otimes e_1+e_2\otimes e_2$ has product $e_1+e_2=1$, and commutes with each algebra factor in the central equation. Thus $A$, and by Section 7 the tensor monad, is separable.

An $A$-module is the pair $(e_1M,e_2M)$: the two idempotents give $M=e_1M\oplus e_2M$, and the scalar action and maps are componentwise. A free module $A\otimes V$ has pair $(V,V)$. The module $(k,0)$ cannot be such a pair, since a common $V$ would have to be both nonzero and zero.

It is the image of multiplication by $e_1$ on $Fk=A$. In the Kleisli category the corresponding map is $p:k\to A\otimes k$, $p(1)=e_1$. Its Kleisli square is $e_1^2=e_1$; its Kleisli identity is $1=e_1+e_2$. The object $(k,p)$ in the completion therefore gives precisely this missing module. The underlying vector-space category is already idempotent complete, but its free-module category is not.

Exercise 8.3 — Enough free modules do not characterize separability (advanced)

Use $k=\mathbb F_p(u^p)\subset L=\mathbb F_p(u)$. For the tensor monad $L\otimes_k-$, prove that its Kleisli/free comparison is an equivalence, although its forgetful functor has no natural Hom retraction.

Solution. Every $L$-module has a basis indexed by some set $I$, so is isomorphic to $L^{(I)}$. The vector space $k^{(I)}$ has free image $L\otimes_k k^{(I)}\simeq L^{(I)}$: the isomorphism takes a tensor with each basis vector to its indicated finitely supported coordinate. Thus every module is free. Proposition 1.1 makes $J$ fully faithful, so it is an equivalence here.

Separable algebra objects and absolute balanced tensor, Exercise 8.3 proves completely that $L$ has degree $p$ over $k$ and is not a separable $k$-algebra: $L\otimes_kL\simeq L[z]/(z^p)$, the central equation forces $ze=0$, and multiplication kills $\operatorname{ann}(z)=Lz^{p-1}$, contradicting product one.

Section 7 therefore says that this tensor monad is not separable. Theorem 3.2 then excludes a natural Hom retraction for its specified forgetful functor. The abstract equivalence with the free category depended on separately chosen module bases; it did not provide the missing natural counit section. This is the stated failure of the converse completion criterion.

Exercise 8.4 — The dual-number module missing from the completion (advanced)

On $\operatorname{Vect}_k$, take $A=k[\epsilon]/(\epsilon^2)$ and $T=A\otimes_k-$. Show that $S=A/(\epsilon)$ is not a retract of any free $T$-module, and hence is absent from the essential image of the completed free comparison.

Solution. A free $T$-module is $A\otimes_k V\simeq A^{(I)}$, choosing a basis $I$ of $V$. Its elements annihilated by $\epsilon$ are exactly $\epsilon A^{(I)}$: in a finitely supported vector of coefficients $a_i+b_i\epsilon$, multiplication by $\epsilon$ is zero exactly when each $a_i=0$.

For an $A$-linear map $f:S\to A^{(I)}$, the generator $1$ of $S$ is annihilated by $\epsilon$, so the entire image lies in $\epsilon A^{(I)}$. Any $A$-linear map $g:A^{(I)}\to S$ kills that submodule. Thus $gf=0$, which cannot be $1_S$. No free retraction exists, even with an infinite-dimensional $V$.

Every object in the Karoubi free category maps to a retract of a free module, since the module category is idempotent complete by Proposition 5.1. It therefore cannot give $S$. In particular this monad cannot be separable, by Theorem 2.1. Concretely its counit at $S$ is the quotient $A\to S$. It has a vector-space section sending $1$ to $1$, but no module section: the image of a possible section is annihilated by $\epsilon$ and therefore maps to zero in $S$. This checks the obstruction on the actual counit.

Freely accessible references

Emily Riehl, Category Theory in Context, author-hosted edition, Section 5.2, for the Kleisli and free-module comparison.
Paul Balmer, Separability and triangulated categories, author manuscript, Section 3, for separable monads and separable forgetful functors.