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adjoint lifting theorem

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Statement

Theorem

(The adjoint lifting theorem). Consider the following commutative square of functors:

π’œ β†’Q ℬ U↓ ↓ V π’ž β†’R π’Ÿ \begin{array}{cccc}\mathcal{A} & \overset{Q}{\to} & \mathcal{B} \\ ^{U}\downarrow & & \downarrow^{V} \\ \mathcal{C} & \underset{R}{\to} & \mathcal{D} \end{array}

and suppose that

Then, if RR has a left adjoint, then QQ also has a left adjoint.

A detailed proof may be found in Sec. 4.5 of Vol. 2 of Borceux (see especially Theorem 4.5.6 on p. 226 and Ex. 4.8.6 on p. 252). Also (Johnstone, prop. 1.1.3) For a sketch of proof, see ahead.

Corollary

If the bottom functor RR of the above square is the identity arrow (so that U=V∘QU=V\circ Q), if UU and VV are monadic, and if π’œ\mathcal{A} has coequalizers of reflexive pairs, then QQ is monadic.

Proof

The adjoint lifting theorem implies the existence of a left adjoint, and the rest is a straightforward application of the monadicity theorem.

Sketch of proof

We may assume the situation of the following diagram (with VQ=RUV Q = R U):

π’ž 𝕋 ⇆QK(?) π’Ÿ π•Š U↓↑ F G↑↓ V π’ž ⇆RL π’Ÿ \begin{array}{cccc}\mathcal{C}^\mathbb{T} & \underoverset{Q}{K(?)}{\leftrightarrows} & \mathcal{D}^\mathbb{S} \\ ^{U}\downarrow \uparrow^{F} & & ^{G}\uparrow\downarrow^{V} \\ \mathcal{C} & \underoverset{R}{L}{\leftrightarrows} & \mathcal{D} \end{array}

where 𝕋=⟨T,Ξ΅:1 π’žβ‡’T,μ⟩\mathbb{T}=\langle T,\varepsilon\colon 1_{\mathcal{C}}\Rightarrow T,\mu \rangle is a monad on π’ž\mathcal{C}, π•Š=⟨S,ΞΆ:1 π’Ÿβ‡’S,η⟩\mathbb{S}=\langle S,\zeta\colon 1_{\mathcal{D}}\Rightarrow S,\eta\rangle is a monad on π’Ÿ\mathcal{D}, UU and VV are the forgetful functors, and FF and GG are the free algebra functors.

Let us write Ο„:FUβ‡’1 π’ž 𝕋\tau\colon F U\Rightarrow 1_{\mathcal{C}^{\mathbb{T}}} for the counit of the adjunction F⊣UF\dashv U and Οƒ:GVβ‡’1 π’Ÿ π•Š\sigma\colon G V\Rightarrow 1_{\mathcal{D}^{\mathbb{S}}} for the counit of the adjunction G⊣VG\dashv V. As usual, we have T=UFT = U F, S=VGS = V G, ΞΌ=UΟ„F\mu=U\tau F, and Ξ·=VΟƒG\eta=V\sigma G.

Finally, let LL be a left adjoint to RR (which exists by assumption), and let Ξ±:1 π’Ÿβ‡’RL\alpha\colon 1_{\mathcal{D}}\Rightarrow R L and Ξ²:LRβ‡’1 π’ž\beta\colon LR\Rightarrow 1_{\mathcal{C}} be the unit and counit (respectively) of the adjunction L⊣RL\dashv R.

We would like to construct a functor K:π’Ÿ π•Šβ†’π’ž 𝕋K\colon \mathcal{D}^{\mathbb{S}}\to \mathcal{C}^{\mathbb{T}}. To get a hint of what KK should look like, let us assume for a moment that such a KK already exists. In this case, we have KG⊣VQ(=RU)K G\dashv V Q(=R U). But FLF L is also left adjoint to RUR U, and by the uniqueness of a left adjoint we must have KG=FLK G = F L (at least up to a natural isomorphism).

From this, we already know how to define KK on free algebras. Also, being a left adjoint, KK preserves in particular all coequalizers. But every π•Š\mathbb{S}-algebra ⟨D,ΞΎ:SDβ†’D⟩\langle D,\xi\colon SD\to D\rangle is the (object part) of a reflexive coequalizer, namely, the canonical presentation

GS(D)⇉GΞΎΟƒ GDG(D)β†’Οƒ ⟨D,ξ⟩=ξ⟨D,ξ⟩ G S(D)\underoverset{G\xi}{\sigma_{G D}}{\rightrightarrows}G(D)\overset{\sigma_{\langle D,\xi\rangle}=\xi}{\rightarrow}\langle D,\xi \rangle

Applying KK and using KG=FLKG=FL, we see that K⟨D,ξ⟩K\langle D,\xi\rangle should be the object part of a reflexive coequalizer in π’ž 𝕋\mathcal{C}^{\mathbb{T}} of the form

FLS(D)⇉FLΞΎ?FL(D)β†’xK⟨D,ξ⟩ F L S(D)\underoverset{F L\xi}{?}{\rightrightarrows}F L(D)\overset{x}{\rightarrow}K\langle D,\xi \rangle

(recall that we assume that π’ž 𝕋\mathcal{C}^{\mathbb{T}} has coequalizers of reflexive pairs).

To eventually define K⟨D,ξ⟩K\langle D,\xi \rangle as a coequalizer (as above), we first need some reasonable guess for the ?-arrow. For this, we will need a lemma.

Lemma

There exists a natural transformation λ:SR⇒RT\lambda\colon S R \Rightarrow R T for which the following diagram of functors and natural transformations is commutative:

R β†’ΞΆR SR ←ηR SSR RΞ΅β†˜ λ↓ Sλ↓ RT ←RΞΌ RTT ←λT SRT \begin{array}{ccccccc} R & \overset{\zeta R}{\to} & S R & & \overset{\eta R}{\leftarrow}& & S S R \\ & ^{R\varepsilon}\searrow & ^{\lambda}\downarrow & & & & ^{S\lambda}\downarrow\\ & & R T & \overset{R\mu}{\leftarrow} & R T T & \overset{\lambda T}{\leftarrow} & S R T \end{array}
Proof

Define Ξ»:=VΟƒQF∘VGRΞ΅\lambda := V\sigma Q F \circ V G R\varepsilon, so that

λ:SR=VGR→VGRΡVGRT=VGRUF=VGVQF→VσQFVQF=RUF=RT. \lambda \colon S R = V G R\overset{V G R\varepsilon}{\to}V G R T = V G R U F = V G V Q F\overset{V\sigma Q F}{\to} V Q F = R U F = R T.

The required commutativity may be verified by using the commutative diagrams in the definitions of a monad and an EM-algebra, naturality, and the triangular identities. For details, see the proof of Lemma 4.5.1 of Borceux, pp. 222-223. (Note that this lemma does not depend neither on the existence of a left adjoint for the bottom horizontal arrow, nor on the existence of coequalizers. Only the commutativity is required.)

We may now return to our task of defining the ?-arrow in the diagram preceding the lemma. We would like to get from FLS(D)F L S(D) to FL(D)F L(D), and for this, we will construct a natural transformation FLS⇒FLF L S\Rightarrow F L in the following way. First, we have

S→SαSRL→λLRTL. S\overset{S\alpha}{\to} S R L \overset{\lambda L}{\to} R T L.

Applying LL and composing with Ξ²TL\beta T L, we get

LS⟢LΞ»L∘LSΞ±LRTLβ†’Ξ²TLTL. LS\overset{L\lambda L\circ LS\alpha}{\longrightarrow}L R T L \overset{\beta TL}{\to} TL.

Applying FF and composing with Ο„FL\tau F L, we finally get

FLS⟢FΞ²TL∘FLΞ»L∘FLSΞ±FTL=FUFLβ†’Ο„FLFL F L S\overset{F\beta T L\circ F L\lambda L\circ F L S\alpha}{\longrightarrow}F T L = F U F L\overset{\tau F L}{\to} F L

Let us call the resulting natural transformation Ο‰\omega, that is,

Ο‰:=Ο„FL∘FΞ²TL∘FLΞ»L∘FLSΞ±. \omega:=\tau F L \circ F\beta T L\circ F L\lambda L\circ F L S\alpha.

Now we take the sought for ?-arrow to be Ο‰ D\omega_D, and define K⟨D,ξ⟩K\langle D,\xi\rangle as the object of some fixed coequalizer of Ο‰ D\omega_D and FLΞΎF L\xi:

FLS(D)⇉FLΞΎΟ‰ DFL(D)β†’xK⟨D,ξ⟩. F L S(D)\underoverset{F L\xi}{\omega_D}{\rightrightarrows}F L(D)\overset{x}{\rightarrow}K\langle D,\xi \rangle.

In order to do this, we must first verify that the parallel arrows above have a common section (since we only assume that π’ž 𝕋\mathcal{C}^{\mathbb{T}} has coequalizers of reflexive pairs). To find a guess for a common section, note that the common section for the parallel pair in the above canonical presentation in π’Ÿ π•Š\mathcal{D}^{\mathbb{S}} is GΞΆ V⟨D,ξ⟩=GΞΆ DG\zeta_{V\langle D,\xi\rangle}=G\zeta_D, and if KK exists, then applying KK gives KGΞΆ D=FLΞΆ DK G\zeta_D=F L\zeta_D. Having this guess, it is now straightforward to verify that FLΞΆ DF L\zeta_D is indeed a common section, as required.

So, we have defined an object function of a would be left adjoint KK. To make it into a functor left adjoint to QQ, we will build a universal arrow from ⟨D,ξ⟩\langle D,\xi\rangle to QQ, whose object part is K⟨D,ξ⟩K\langle D,\xi\rangle (Theorem IV.1.2(ii) of Categories Work).

To get an arrow ⟨D,ΞΎβŸ©β†’QK⟨D,ξ⟩\langle D,\xi\rangle\to Q K\langle D,\xi\rangle , suppose for a moment that we have a natural transformation Ο†:Gβ‡’QFL\varphi\colon G\Rightarrow Q F L such that Ο†βˆ˜ΟƒG=QΟ‰βˆ˜Ο†S\varphi\circ \sigma G = Q\omega\circ \varphi S. Then the left square in the following diagram commutes:

GS(D) ⇉GΞΎΟƒ GD G(D) β†’Οƒ ⟨D,ξ⟩=ΞΎ ⟨D,ξ⟩ Ο† SD↓ Ο† D↓ χ↓ QFLSD ⇉QFLΞΎQΟ‰ D QFLD β†’Qx QK⟨D,ξ⟩ \begin{array}{ccccc}G S(D)& \underoverset{G\xi}{\sigma_{G D}}{\rightrightarrows} & G(D) &\overset{\sigma_{\langle D,\xi\rangle}=\xi}{\rightarrow}&\langle D,\xi \rangle \\ ^{\varphi_{S D}}\downarrow && ^{\varphi_D}\downarrow && ^{\chi}\downarrow\\ Q F L S D &\underoverset{Q F L\xi}{Q\omega_D}{\rightrightarrows} & Q F L D &\overset{Q x}{\rightarrow}&Q K\langle D,\xi\rangle \end{array}

Since both rows are forks, it follows that Qxβˆ˜Ο† DQ x\circ \varphi_D has the same composition with the arrows of the upper
parallel pair, and hence there exists a unique arrow Ο‡:⟨D,ΞΎβŸ©β†’QK⟨D,ξ⟩\chi\colon \langle D,\xi\rangle\to Q K\langle D,\xi\rangle making the right square commutative (recall that the upper row is a coequalizer).

It is now possible to prove that the pair ⟨K⟨D,ξ⟩,Ο‡βŸ©\langle K\langle D,\xi\rangle,\chi \rangle is a universal arrow from ⟨D,ξ⟩\langle D,\xi\rangle to QQ, showing that KK is indeed (the object function) of a left adjoint (for details, see the proof of Theorem 4.5.6, pp. 226-227 in Borceux).

But we still have to prove the existence of a natural transformation Ο†:Gβ‡’QFL\varphi\colon G\Rightarrow Q F L such that Ο†βˆ˜ΟƒG=QΟ‰βˆ˜Ο†S\varphi\circ \sigma G = Q\omega\circ \varphi S. For this, we define Ο†:=ΟƒQFL∘GRΞ΅L∘GΞ±\varphi:=\sigma Q F L \circ G R\varepsilon L \circ G\alpha. Since VV is faithful, to prove the required property of Ο†\varphi, it is enough to prove that VΟ†βˆ˜VΟƒG=VQΟ‰βˆ˜VΟ†SV\varphi \circ V\sigma G = V Q\omega\circ V\varphi S, and this is a long, yet straightforward, computation (noting that VΟ†=Ξ»L∘VGΞ±V\varphi = \lambda L \circ VG\alpha and using the commutative diagram from Lemma 1; see Lemma 4.5.3, p. 224 of Borceux).

Examples

Forgetful functors between varieties of algebras

Since varieties of algebras are cocomplete and monadic over Set\mathbf{Set}, the corollary implies that forgetful functors between varieties of algebras (e.g., the forgetful functor Rng→Ab\mathbf{Rng}\to\mathbf{Ab}) are monadic.

Sufficient conditions for cocompleteness of monadic categories

Let π’₯\mathcal{J} be an arbitrary category, and consider the commutative diagram

π’œ β†’Ξ” π’œ π’₯ U↓ ↓ U π’₯ π’ž β†’Ξ” π’ž π’₯ \begin{array}{ccccc}\mathcal{A} & \overset{\Delta}{\to} & \mathcal{A}^\mathcal{J} \\ ^{U}\downarrow & & \downarrow^{U^{\mathcal{J}}} \\ \mathcal{C} & \underset{\Delta}{\to} & \mathcal{C}^\mathcal{J} \end{array}

where UU is monadic, Ξ”\Delta is the diagonal functor and U π’₯=Uβˆ˜βˆ’U^{\mathcal{J}}=U\circ -. If FF is left adjoint
to UU, then F π’₯F^\mathcal{J} is left adjoint to U π’₯U^{\mathcal{J}} (using the unit and counit of the original adjunction, one can construct appropriate natural transformations that satisfy the triangular identities, see, e.g., p. 119 of Categories Work). Also, the conditions of the monadicity theorem for UU imply those for U π’₯U^\mathcal{J} (basically because the definition of a split fork involves only compositions and identities, and because natural transformations are composed componentwise).

Now, if π’ž\mathcal{C} is π’₯\mathcal{J}-cocomplete (so that the bottom horizontal functor has a left adjoint) and π’œ\mathcal{A} has coequalizers of reflexive pairs, then the adjoint lifting theorem implies that π’œ\mathcal{A} is π’₯\mathcal{J}-cocomplete. In particular, if π’œ\mathcal{A} has coequalizers of reflexive pairs and π’ž\mathcal{C} is small-cocomplete, then π’œ\mathcal{A} is small cocomplete.

References

Section 4.5 of volume 2 of

  • Peter Johnstone, Adjoint lifting theorems for categories of algebras , Bull. London Math. Soc., 7 (1975), 294-297.

Section A1.1 of

Revised on October 12, 2012 22:41:51 by Mike Shulman (192.16.204.218)