Adjunctions Units and Counits

Starting with an adjoint equivalence, forget only the assertion that $\eta$ and $\epsilon$ are invertible. The remaining functors, natural transformations, and triangle identities are an adjunction, while the forgotten assertion still says its unit and counit are natural isomorphisms.

Conversely, an adjunction with invertible unit and counit has exactly the functors, natural isomorphisms, and triangle identities required by [F1], so it is an adjoint equivalence.

Finally, [F2] equips any equivalence with the data in step 1.1, proving the consequence.

Expanding the first composite gives $(u^{\mathrm{♭}}{)}^{\mathrm{♯}}={\epsilon }_d\circ FG(u)\circ F({\eta }_c)$.

Naturality of $\epsilon$ at $u$ gives ${\epsilon }_d\circ FG(u)=u\circ {\epsilon }_{Fc}$.

Expanding the second composite gives $(v^{\mathrm{♯}}{)}^{\mathrm{♭}}=G({\epsilon }_d)\circ GF(v)\circ {\eta }_c$.

Naturality of $\eta$ at $v$ gives $GF(v)\circ {\eta }_c={\eta }_{Gd}\circ v$.

Substituting step 1.2 into step 1.1 and applying the first triangle identity yields $(u^{\mathrm{♭}}{)}^{\mathrm{♯}}=u\circ {\epsilon }_{Fc}\circ F({\eta }_c)=u$.

Substituting step 1.4 into step 1.3 and applying the second triangle identity yields $(v^{\mathrm{♯}}{)}^{\mathrm{♭}}=G({\epsilon }_d)\circ {\eta }_{Gd}\circ v=v$.

Given an adjunction, [F1] makes the displayed hom-collections sets, and [L1] shows that ${\mathrm{\Phi }}_{c,d}$ and ${\mathrm{\Psi }}_{c,d}$ are inverse bijections.

For $a:c^{\mathrm{\prime }}\to c$, $b:d\to d^{\mathrm{\prime }}$, and $u:Fc\to d$, functoriality and naturality of $\eta$ give ${\mathrm{\Phi }}_{c^{\mathrm{\prime }},d^{\mathrm{\prime }}}(b\circ u\circ F(a))=G(b)\circ {\mathrm{\Phi }}_{c,d}(u)\circ a$; this is naturality in both variables in the sense of [F2].

Conversely, let $\mathrm{\Phi }$ be a natural family of bijections with inverse $\mathrm{\Psi }$. Define ${\eta }_c:={\mathrm{\Phi }}_{c,Fc}(1_{Fc})$ and ${\epsilon }_d:={\mathrm{\Psi }}_{Gd,d}(1_{Gd})$.

Naturality of $\mathrm{\Phi }$ in $c$ makes $\eta$ natural, and naturality of $\mathrm{\Psi }$ in $d$ makes $\epsilon$ natural.

Naturality in $d$ and $c$, respectively, gives ${\mathrm{\Phi }}_{c,d}(u)=G(u)\circ {\eta }_c$ and ${\mathrm{\Psi }}_{c,d}(v)={\epsilon }_d\circ F(v)$.

Since $\mathrm{\Psi }\mathrm{\Phi }$ fixes $1_{Fc}$, step 2.2 gives ${\epsilon }_{Fc}\circ F({\eta }_c)=1_{Fc}$; since $\mathrm{\Phi }\mathrm{\Psi }$ fixes $1_{Gd}$, it gives $G({\epsilon }_d)\circ {\eta }_{Gd}=1_{Gd}$.

Thus the data of steps 1.3 and 2.1 satisfy both triangle identities and define an adjunction. Any unit and counit inducing $\mathrm{\Phi }$ must be the transposes of the two identity morphisms, so step 1.3 also proves uniqueness.

If $b\circ u=u^{\mathrm{\prime }}\circ F(a)$, apply transposition. By [L1], the transpose of the left side is $G(b)\circ u^{\mathrm{♭}}$, while the transpose of the right side is $(u^{\mathrm{\prime }}{)}^{\mathrm{♭}}\circ a$, so the transposed square commutes.

Conversely, if the transposed square commutes, apply the inverse bijection to its two sides. The two inverse images are $b\circ u$ and $u^{\mathrm{\prime }}\circ F(a)$ by [L1], so the original square commutes.

Repeating steps 1.1 and 1.2 with the inverse bijections proves the analogous assertion for inverse transposition.

Fix $c$ and a morphism $f:c\to Gd$. Its inverse transpose $f^{\mathrm{♯}}:Fc\to d$ satisfies $G(f^{\mathrm{♯}})\circ {\eta }_c=f$, and it is the unique morphism with this property because transposition is injective.

Dually, for $u:Fc\to d$, its transpose $u^{\mathrm{♭}}:c\to Gd$ is the unique morphism satisfying ${\epsilon }_d\circ F(u^{\mathrm{♭}})=u$.

Thus $(Fc,{\eta }_c)$ has the universal property in [F1], and [F2] makes it initial in $(c\downarrow G)$.

Hence $(Gd,{\epsilon }_d)$ is universal from $F$ to $d$ and terminal in $(F\downarrow d)$. The argument used individual morphisms and uniqueness only, so it imposed no set-size condition on any hom-class.

Descriptions 1 and 2 determine one another by [L2], including the recovery formulas obtained by transposing identity morphisms.

Description 1 gives descriptions 3 and 4 by [L3].

Conversely, from description 3, put $u^{\mathrm{♭}}:=G(u)\circ {\eta }_c$ for $u:Fc\to d$; the unique-factorisation property of $(Fc,{\eta }_c)$ says exactly that this is a bijection onto the morphisms $c\to Gd$, with inverse the unique factorisation. Define ${\epsilon }_d:FGd\to d$ as the unique morphism with $G({\epsilon }_d)\circ {\eta }_{Gd}=1_{Gd}$, which is the second triangle identity. For $g:d\to d^{\mathrm{\prime }}$ the two morphisms $g\circ {\epsilon }_d$ and ${\epsilon }_{d^{\mathrm{\prime }}}\circ FG(g)$ have the same factorisation datum, since $G(g)G({\epsilon }_d){\eta }_{Gd}=G(g)$ and $G({\epsilon }_{d^{\mathrm{\prime }}})GFG(g){\eta }_{Gd}=G({\epsilon }_{d^{\mathrm{\prime }}}){\eta }_{G{d}^{\mathrm{\prime }}}G(g)=G(g)$; uniqueness makes $\epsilon$ natural. Likewise $G({\epsilon }_{Fc})GF({\eta }_c){\eta }_c=G({\epsilon }_{Fc}){\eta }_{GFc}{\eta }_c={\eta }_c=G(1_{Fc}){\eta }_c$, so uniqueness gives the first triangle identity ${\epsilon }_{Fc}F({\eta }_c)=1_{Fc}$. This recovers a unit and counit satisfying [L1].

The dual construction from description 4 defines ${\eta }_c:c\to GFc$ as the unique morphism with ${\epsilon }_{Fc}\circ F({\eta }_c)=1_{Fc}$, proves its naturality from terminality of $(Gd,{\epsilon }_d)$ in $(F\downarrow d)$ by the argument of step 1.3 read in the opposite categories, and yields the second triangle identity the same way. It therefore gives the same unit-counit data.

Steps 1.1 through 2.1 establish all implications. Since [L3] is expressed by unique individual factorisations, descriptions 1, 3, and 4 remain meaningful without local smallness, whereas [L2] explicitly uses hom-sets.

If a left adjoint $F$ is supplied, [L1] gives the required chosen initial object $(Fc,{\eta }_c)$ for every $c$.

Conversely, suppose such initial objects are supplied. For $a:c\to c^{\mathrm{\prime }}$, initiality gives a unique $F(a):Fc\to F{c}^{\mathrm{\prime }}$ satisfying $G(F(a))\circ {\eta }_c={\eta }_{c^{\mathrm{\prime }}}\circ a$.

The identity $1_{Fc}$ satisfies the defining equation for $F(1_c)$, so uniqueness gives $F(1_c)=1_{Fc}$.

If $a:c\to c^{\mathrm{\prime }}$ and $b:c^{\mathrm{\prime }}\to c^{\mathrm{\prime }\mathrm{\prime }}$, then $F(b)F(a)$ satisfies the defining equation for $F(ba)$; uniqueness gives $F(ba)=F(b)F(a)$. Thus $F$ is a functor and $\eta$ is natural.

For every $f:c\to Gd$, initiality supplies a unique $\widehat{f}:Fc\to d$ with $G(\widehat{f}){\eta }_c=f$. Steps 2.1 and 2.2 make $F$ a functor and $\eta$ natural, so $(Fc,{\eta }_c{)}_c$ is a natural family of universal arrows from each $c$ to $G$; by [L2] that data is an adjunction, so $F⊣G$.

Any functor action compatible with the chosen initial objects must satisfy the equation in step 1.2 and is therefore equal to this one. Passing to opposite categories proves the terminal-object criterion for right adjoints.

For $a:c\to c^{\mathrm{\prime }}$, apply [F1] to ${\eta }_{c^{\mathrm{\prime }}}a:c\to GF{c}^{\mathrm{\prime }}$ and define $F(a)$ as the unique morphism satisfying $G(F(a)){\eta }_c={\eta }_{c^{\mathrm{\prime }}}a$.

The uniqueness clause in [F1] gives $F(1_c)=1_{Fc}$ and $F(ba)=F(b)F(a)$, because the proposed right sides satisfy the same defining equations.

Hence $F$ is a functor and the equations in step 1.1 say exactly that $\eta$ is natural. The same universal factorisations give the adjunction $F⊣G$ by [L1].

Any other compatible functor structure would have to satisfy the defining equation in step 1.1, so [F1] makes it identical to this one on every morphism.

For $h:d\to d^{\mathrm{\prime }}$, postcomposition by $h$ gives a natural transformation $\mathcal{D}(F(-),d)\Rightarrow \mathcal{D}(F(-),d^{\mathrm{\prime }})$. Transport it through ${\theta }_d$ and ${\theta }_{d^{\mathrm{\prime }}}$; [F2] supplies a unique morphism $G(h):Gd\to G{d}^{\mathrm{\prime }}$ representing the result.

Postcomposition by an identity is the identity transformation, so Yoneda uniqueness gives $G(1_d)=1_{Gd}$.

Postcomposition by $kh$ is the composite of postcomposition by $h$ and by $k$, so Yoneda uniqueness gives $G(kh)=G(k)G(h)$. Thus $G$ is a functor.

The definition in step 1.1 makes $\theta$ natural in $d$; it was natural in $c$ by hypothesis. Therefore [L1] gives $F⊣G$.

If another functor structure made every ${\theta }_d$ natural, its value on $h$ would induce the same transported natural transformation, so [F2] would force it to equal $G(h)$. The supplied object assignment also shows that no class-sized selection was made in the proof.

By initiality of $(Fc,{\eta }_c)$ there is a unique ${\alpha }_c:Fc\to F^{\mathrm{\prime }}c$ with $G({\alpha }_c){\eta }_c={\eta }_c^{\mathrm{\prime }}$; reversing the roles gives a unique ${\beta }_c:F^{\mathrm{\prime }}c\to Fc$ with $G({\beta }_c){\eta }_c^{\mathrm{\prime }}={\eta }_c$.

The composites ${\beta }_c{\alpha }_c$ and $1_{Fc}$ both carry ${\eta }_c$ to itself, so initiality gives ${\beta }_c{\alpha }_c=1_{Fc}$; similarly ${\alpha }_c{\beta }_c=1_{F^{\mathrm{\prime }}c}$.

Let $a:c\to c^{\mathrm{\prime }}$. The object $(F^{\mathrm{\prime }}{c}^{\mathrm{\prime }},{\eta }_{c^{\mathrm{\prime }}}^{\mathrm{\prime }}\circ a)$ lies in $(c\downarrow G)$, so by initiality of $(Fc,{\eta }_c)$ exactly one morphism $Fc\to F^{\mathrm{\prime }}{c}^{\mathrm{\prime }}$ composes with ${\eta }_c$ to ${\eta }_{c^{\mathrm{\prime }}}^{\mathrm{\prime }}a$. Both candidates do: $G(F^{\mathrm{\prime }}a{\alpha }_c){\eta }_c=G(F^{\mathrm{\prime }}a){\eta }_c^{\mathrm{\prime }}={\eta }_{c^{\mathrm{\prime }}}^{\mathrm{\prime }}a$ by naturality of ${\eta }^{\mathrm{\prime }}$, and $G({\alpha }_{c^{\mathrm{\prime }}}Fa){\eta }_c=G({\alpha }_{c^{\mathrm{\prime }}}){\eta }_{c^{\mathrm{\prime }}}a={\eta }_{c^{\mathrm{\prime }}}^{\mathrm{\prime }}a$ by naturality of $\eta$. Hence $F^{\mathrm{\prime }}a{\alpha }_c={\alpha }_{c^{\mathrm{\prime }}}Fa$ and $\alpha$ is natural. This uses only initiality and naturality of the units, so no local smallness is required.

Any natural transformation compatible with the units has components satisfying the uniqueness condition in step 1.1 and hence equals $\alpha$. For counit compatibility, both ${\epsilon }_d$ and ${\epsilon }_d^{\mathrm{\prime }}\circ {\alpha }_{Gd}$ are morphisms $FGd\to d$, and initiality of $(FGd,{\eta }_{Gd})$ determines such a morphism by its composite: $G({\epsilon }_d){\eta }_{Gd}=1_{Gd}$ by the triangle identity of [L1], while $G({\epsilon }_d^{\mathrm{\prime }})G({\alpha }_{Gd}){\eta }_{Gd}=G({\epsilon }_d^{\mathrm{\prime }}){\eta }_{Gd}^{\mathrm{\prime }}=1_{Gd}$ by step 1.1 and the triangle identity for $F^{\mathrm{\prime }}⊣G$. So ${\epsilon }_d^{\mathrm{\prime }}\circ {\alpha }_{Gd}={\epsilon }_d$. Passing to opposite categories proves the dual assertion.

Whiskering gives $G{\eta }^{\mathrm{\prime }}F:GF\Rightarrow G{G}^{\mathrm{\prime }}{F}^{\mathrm{\prime }}F$ and $F^{\mathrm{\prime }}\epsilon G^{\mathrm{\prime }}:F^{\mathrm{\prime }}FG{G}^{\mathrm{\prime }}\Rightarrow F^{\mathrm{\prime }}{G}^{\mathrm{\prime }}$, so the displayed composites have the required types; they are natural because whiskering and vertical composition preserve naturality.

Expanding the first triangle composite for $F^{\mathrm{\prime }}F$ gives $(\bar{\epsilon }{F}^{\mathrm{\prime }}F)\circ (F^{\mathrm{\prime }}F\bar{\eta })=({\epsilon }^{\mathrm{\prime }}{F}^{\mathrm{\prime }}F)\circ (F^{\mathrm{\prime }}\epsilon G^{\mathrm{\prime }}{F}^{\mathrm{\prime }}F)\circ (F^{\mathrm{\prime }}FG{\eta }^{\mathrm{\prime }}F)\circ (F^{\mathrm{\prime }}F\eta )$. Interchange rewrites the two middle factors as $F^{\mathrm{\prime }}[(\epsilon G^{\mathrm{\prime }}{F}^{\mathrm{\prime }}F)\circ (FG{\eta }^{\mathrm{\prime }}F)]$, and naturality of $\epsilon$ at the component ${\eta }_{Fc}^{\mathrm{\prime }}$ turns that bracket into $({\eta }^{\mathrm{\prime }}F)\circ (\epsilon F)$.

Expanding the second triangle composite for $G{G}^{\mathrm{\prime }}$ gives $(G{G}^{\mathrm{\prime }}\bar{\epsilon })\circ (\bar{\eta }G{G}^{\mathrm{\prime }})=(G{G}^{\mathrm{\prime }}{\epsilon }^{\mathrm{\prime }})\circ (G{G}^{\mathrm{\prime }}{F}^{\mathrm{\prime }}\epsilon G^{\mathrm{\prime }})\circ (G{\eta }^{\mathrm{\prime }}FG{G}^{\mathrm{\prime }})\circ (\eta G{G}^{\mathrm{\prime }})$. Interchange rewrites the two middle factors as $G[(G^{\mathrm{\prime }}{F}^{\mathrm{\prime }}\epsilon G^{\mathrm{\prime }})\circ ({\eta }^{\mathrm{\prime }}FG{G}^{\mathrm{\prime }})]$, and naturality of ${\eta }^{\mathrm{\prime }}$ at the component ${\epsilon }_{G^{\mathrm{\prime }}d}$ turns that bracket into $({\eta }^{\mathrm{\prime }}{G}^{\mathrm{\prime }})\circ (\epsilon G^{\mathrm{\prime }})$. The composite becomes $G[(G^{\mathrm{\prime }}{\epsilon }^{\mathrm{\prime }})\circ ({\eta }^{\mathrm{\prime }}{G}^{\mathrm{\prime }})]\circ [(G\epsilon )\circ (\eta G)]G^{\mathrm{\prime }}$, which by the two second triangle identities $G\epsilon \circ \eta G=1_G$ and $G^{\mathrm{\prime }}{\epsilon }^{\mathrm{\prime }}\circ {\eta }^{\mathrm{\prime }}{G}^{\mathrm{\prime }}=1_{G^{\mathrm{\prime }}}$ is $1_{G{G}^{\mathrm{\prime }}}$.

By step 2.1 the first composite becomes $[({\epsilon }^{\mathrm{\prime }}{F}^{\mathrm{\prime }})\circ (F^{\mathrm{\prime }}{\eta }^{\mathrm{\prime }})]F\circ F^{\mathrm{\prime }}[(\epsilon F)\circ (F\eta )]$, which by the two first triangle identities $\epsilon F\circ F\eta =1_F$ and ${\epsilon }^{\mathrm{\prime }}{F}^{\mathrm{\prime }}\circ F^{\mathrm{\prime }}{\eta }^{\mathrm{\prime }}=1_{F^{\mathrm{\prime }}}$ is $1_{F^{\mathrm{\prime }}F}$.

Thus $\bar{\eta }$ and $\bar{\epsilon }$ satisfy both triangle identities, and [L1] gives $F^{\mathrm{\prime }}F⊣G{G}^{\mathrm{\prime }}$.

Postcomposition sends $H:\mathcal{J}\to \mathcal{C}$ to $FH$ and $K:\mathcal{J}\to \mathcal{D}$ to $GK$. Whiskering $\eta$ and $\epsilon$ gives unit components $\eta H:H\Rightarrow GFH$ and counit components $\epsilon K:FGK\Rightarrow K$.

Precomposition sends $H:\mathcal{C}\to \mathcal{E}$ to $HG$ and $K:\mathcal{D}\to \mathcal{E}$ to $KF$. Whiskering now gives the unit $H\eta :H\Rightarrow HGF$ and counit $K\epsilon :KFG\Rightarrow K$, so $G^{\ast }⊣F^{\ast }$.

The two triangle identities hold at every object of $\mathcal{J}$ by [L1], hence hold as equalities of natural transformations. Therefore $F_{\ast }⊣G_{\ast }$.

Again the triangle identities are the images under $H$ and $K$ of those in [L1]. The size assertion follows from [F2]; the componentwise unit-counit construction itself uses only legitimate functors and natural transformations.

Whiskering shows that the three factors defining ${\alpha }^{\mathrm{♭}}$ have successive types $HG\to G^{\mathrm{\prime }}{F}^{\mathrm{\prime }}HG\to G^{\mathrm{\prime }}KFG\to G^{\mathrm{\prime }}K$; the three factors defining ${\beta }^{\mathrm{♯}}$ have successive types $F^{\mathrm{\prime }}H\to F^{\mathrm{\prime }}HGF\to F^{\mathrm{\prime }}{G}^{\mathrm{\prime }}KF\to KF$.

Each of those six factors is a whiskering of one of ${\eta }^{\mathrm{\prime }}$, $\alpha$, $\epsilon$, ${\epsilon }^{\mathrm{\prime }}$, $\beta$, $\eta$, every one of which is natural by [L1] and the hypothesis on $\alpha$ and $\beta$; so each factor is natural by [L2], and the two vertical composites are natural by [L3]. Only the unit–counit data enters, so the argument applies whether or not the four categories are locally small.

When the two adjunctions coincide and $H,K$ are identity functors, the middle factor $G{1}_{F}G$ is the identity transformation of $GFG$, so the mate of $1_F$ is $(G\epsilon )\circ (\eta G)$, which is $1_G$ by the triangle identity of [L1]. For a typed vertical or horizontal pasting, expand the displayed formulas; interchange identifies the expansion with the corresponding pasting of the mates, with the order fixed by the types.

Substitute ${\alpha }^{\mathrm{♭}}$ into the formula for $(-{)}^{\mathrm{♯}}$. Interchange moves the two unit-counit pairs together, and the triangle identities cancel both pairs, leaving $\alpha$.

Substituting ${\beta }^{\mathrm{♯}}$ into the formula for $(-{)}^{\mathrm{♭}}$ gives the dual cancellation and leaves $\beta$. Thus the constructions are inverse.

Hence the formulas give the asserted bijection, carry $1_F$ to $1_G$ in the coinciding-adjunction case, and preserve both forms of compatible pasting.

If ${\eta }_c$ is invertible, then $F({\eta }_c)$ is invertible, and the first triangle identity makes ${\epsilon }_{Fc}=F({\eta }_c{)}^{-1}$; hence $Fc\in {\mathcal{D}}_{\epsilon }$.

Dually, if ${\epsilon }_d$ is invertible, then the second triangle identity makes ${\eta }_{Gd}=G({\epsilon }_d{)}^{-1}$; hence $Gd\in {\mathcal{C}}_{\eta }$.

By [F1], $F$ and $G$ therefore restrict to functors between the two full subcategories, and the restricted unit and counit remain natural.

Every component of the restricted unit and counit is an isomorphism by definition, so [F2] makes the restrictions quasi-inverse equivalences; the inherited triangle identities make the equivalence adjoint.

If $G$ is faithful and $r{\epsilon }_d=s{\epsilon }_d$, then $G(r)G({\epsilon }_d)=G(s)G({\epsilon }_d)$; composing with ${\eta }_{Gd}$ and using the triangle identity gives $G(r)=G(s)$, hence $r=s$. Thus each ${\epsilon }_d$ is epic.

Conversely, if every ${\epsilon }_d$ is epic and $G(r)=G(s)$, naturality gives $r{\epsilon }_d={\epsilon }_{e}FG(r)={\epsilon }_{e}FG(s)=s{\epsilon }_d$, so $r=s$. Hence $G$ is faithful.

If $G$ is full, lift ${\eta }_{Gd}:Gd\to GFGd$ to $q_d:d\to FGd$ with $G(q_d)={\eta }_{Gd}$. Naturality of $\epsilon$ and the first triangle identity give $q_d{\epsilon }_d={\epsilon }_{FGd}FG(q_d)={\epsilon }_{FGd}F({\eta }_{Gd})=1_{FGd}$, so ${\epsilon }_d$ is split monic.