#adjoint functors

about non-adjoint equivalences between categories was wrong, because I didn’t make myself check that an “obvious-looking” unit map was actually a natural transformation.

Anyways, I spent last night re-convincing myself that non-adjoint equivalences still exist.

I ended up checking the following (identifying a group with its delooping, i.e. the one-object category carrying that group as the automorphism group of the unique object):

Any equivalence \(F : H \leftrightarrows H : G\) of a group with itself comprises two automorphisms \(F, G\), such that \(F G\) and \(G F\) are inner. The unit and counit are the group elements \(g_{\rho}\) such that \(GF(k) = g_{\rho} k g_{\rho}^{-1}\) and  \(g_{\sigma}\) such that \(FG(k) = g_{\sigma}^{-1} k g_{\sigma}\) for any \(k \in H\).

Any equivalence of \(H\) with itself where \(F\) and \(G\) are themselves also inner is an adjoint equivalence.

If \(H\) has trivial center, then any equivalence of \(H\) with itself is an adjoint equivalence.

To obtain a non-adjoint equivalence, we therefore need a group \(H\) with nontrivial center and nontrivial outer automorphisms, such that we can pick two whose products are inner.

So take \(H = K\) the Klein 4-group. This is a product of abelian groups, so abelian, so is its own center. In fact, it’s \(\mathbb{Z}/2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z}\), so let \(F = G\) the automorphism which interchanges coordinates. \(FG = GF = \operatorname{id}_{K}\), which is given by conjugation by any element.

One of the triangle equalities stipulates that \(F(g_{\rho}) = g_{\sigma}^{-1}\). We can pick \(g_{\rho}\) and \(g_{\sigma}\) to break this. For example, let \(g_{\rho}\) be \((1,1)\) and let \(g_{\sigma}\) be \((0,1)\).

Lately I’ve been thinking about the definability of adjunctions between definable categories. I found myself using the unit-counit formulation of adjointness fairly often, because it makes some parts of the adjunction very explicit. But I realized I’ve never carried out the exercise of showing the two formulations are equivalent, so here it is.

Definition 1. A pair of functors \(F : \mathbf{C} \rightleftarrows \mathbf{D} : G\) is a hom-set adjunction if there exists a family of bijections \[\big\{\phi_{X,Y} : \operatorname{Hom}_{\mathbf{D}}(FX,Y) \overset{\sim}{\to} \operatorname{Hom}_{\mathbf{C}}(X,GY)\big\},\] natural in \(X\) and \(Y\).

Definition 2. A pair of functors \(F: \mathbf{C} \rightleftarrows \mathbf{D} : G\) is a unit-counit adjunction if there exist natural transformations \[\eta : 1_{\mathbf{C}} \to GF \text{ and } \epsilon : FG \to 1_{\mathbf{D}}\] such that the maps \[F \overset{F \eta}{\to} FGF \overset{\epsilon F}{\to} F\] and \[G \overset{\eta G}{\to} GFG \overset{G \epsilon}{\to} G\] are the identities \(\operatorname{id}_{F}\) and \(\operatorname{id}_{G}\), respectively. These are called the triangle identities for \(\eta\) and \(\epsilon\).

Theorem 1. \(F : \mathbf{C} \rightleftarrows \mathbf{D} : G\) are hom-set adjoint if and only if they are unit-counit adjoint.

Proof. Suppose first that \(F\) and \(G\) are hom-set adjoint. Obtain the components of the unit map \(\{\eta_X\}_{X \in \mathbf{C}}\) by setting \[\eta_X \overset{\operatorname{df}}{=}\overline{\operatorname{id}_{FX}} \text{ \(= \phi_{X,FX}(\operatorname{id}_{FX} : FX \to FX)\)}\] and the components of the counit map \(\{\epsilon_Y\}_{Y \in \mathbf{D}}\) by setting \[\epsilon_Y \overset{\operatorname{df}}{=}\overline{\operatorname{id}_{GY}} \text{ \(= \phi_{GY,Y}^{-1}(\operatorname{id}_{GY} : GY \to GY)\)}.\] To check naturality, let \(X_1 \overset{f}{\to} X_2\) be a map in \(\mathbf{C}\). Then form the commutative diagram

so that chasing \(\overline{\operatorname{id}_{FX_2}}\) yields

Similarly, forming the commutative diagram

 and chasing \(\overline{\operatorname{id}_{FX_1}}\) yields

so that \(\phi^{-1}_{X_1, FX_2}(\overline{\operatorname{id}_{FX_2}} \circ f) = \phi^{-1}_{X_1, FX_2}(GFf \circ \overline{\operatorname{id}_{FX_1}}) \implies \overline{\operatorname{id}_{FX_2}} \circ f = GFf \circ \overline{\operatorname{id}_{FX_1}},\) hence \(\{\eta_X\}_{X \in \mathbf{C}}\) is natural in \(X\) .

We can argue similarly to show the naturality of \(\epsilon_Y\). Let \(Y_1 \overset{f}{\to} Y_2\) be a map in \(\mathbf{D}\). Then form the commutative diagram

and chase \(\operatorname{id}_{GY_1}\) to obtain

Now, forming the commutative diagram

and chasing \(\operatorname{id}_{GY_2}\) yields

so that by the bijectivity of \(\phi_{GY_1, Y_2}\), we have the equality \(f \circ \overline{\operatorname{id}_{GY_1}} = \overline{\operatorname{id}_{GY_2}} \circ FGf\). Hence \(\{\epsilon_Y\}_{Y \in \mathbf{D}}\) is natural in \(Y\).

It suffices to check the triangle identities componentwise. We require: \[\big( FX \overset{F \overline{\operatorname{id}_{FX}}}{\to} FGFX \overset{\overline{\operatorname{id}_{GFX}}}{\to} FX \big) = FX \overset{\operatorname{id}_{FX}}{\to} FX.\] This follows from chasing \(\operatorname{id}_{GFX}\) around the square

Similarly, we require for each \(Y \in \mathbf{D}\) that \[\big(GY \overset{\overline{\operatorname{id}_{FGY}}}{\to} GFGY \overset{G \overline{\operatorname{id}_{GY}}}{\to} GY \big) = GY \overset{\operatorname{id}_{GY}}{\to} GY.\] This follows from chasing \(\operatorname{id}_{FGY}\) around the square (where we have inverted the \(\phi\)’s)

which completes this direction of the proof.

Now suppose that \(F\) and \(G\) are unit-counit adjoint. If we have a map \(FX \overset{f}{\to} Y\), if we did have the hom-set adjunction, we must have a commutative diagram

so that chasing \(\overline{\operatorname{id}_{FX}}\) yields

Similarly, if we have a map \(X \overset{f}{\to} GY\), the hom-set adjunction must yield the commutative diagram

so that chasing \(\operatorname{id}_{GY}\) yields

Hence, \(\phi_{X,Y}\) must be given by \(f \mapsto Gf \circ \overline{\operatorname{id}_{FX}}\), and its inverse \(\phi^{-1}_{X,Y}\) must be given by \(f \mapsto \overline{\operatorname{id}_{GY}} \circ Ff\).

To see that they are inverse, consider \(\phi^{-1}_{X,Y} \circ \phi_{X,Y}(f)\), which is given by \[f \mapsto Gf \circ \overline{\operatorname{id}_{FX}} \mapsto \overline{\operatorname{id}_{GY}} \circ F \left(Gf \circ \overline{\operatorname{id}_{FX}}\right).\] By the naturality of the counit \(\epsilon\), the square 

commutes, so that \[\overline{\operatorname{id}_{GY}} \circ FGf \circ F \overline{\operatorname{id}_{FX}} = f \circ \overline{\operatorname{id}_{GFX}} \circ F \overline{\operatorname{id}_{FX}} = f,\]whence the triangle identities for the unit-counit adjunction. Similarly, if \(f : X \to GY\), then \(\phi \circ \phi^{-1}(f)\) is given by \[f \mapsto \overline{\operatorname{id}_{GY}} \circ Ff \mapsto G \left(\overline{\operatorname{id}_{GY}} \circ Ff \right) \circ \overline{\operatorname{id}_{FX}}\] and by the naturality of the unit \(\eta\), the square 

commutes, so that \[G \overline{\operatorname{id}_{GY}} \circ GFf \circ \overline{\operatorname{id}_{FX}} = G \overline{\operatorname{id}_{GT}} \circ \overline{\operatorname{id}_{FGY}} \circ f = f,\]whence the triangle identities for the unit-counit adjunction. Hence, \(\phi_{X,Y}\) and \(\phi^{-1}_{X,Y}\) invert each other, as their nomenclature suggests.

Now naturality. A map \(X_1 \overset{f}{\to} X_2\) yields

which at the level of elements yields (by how we’ve defined \(\phi\) and \(\phi^{-1}\)) the identity 

so that \(\phi\) is natural in \(X\). Similarly, a map \(Y_1 \overset{f}{\to} Y_2\) yields

which at the level of elements yields (by how we’ve defined \(\phi\) and \(\phi^{-1}\)) the identity

Suppose we have categories $C$ and $D$ and a functor $C \overset{R}{\to} D$. Suppose $R$ has the property that for all $d \in |D|$, $d$ has a reflection across $R$, i.e. an initial object in $d \downarrow R$, which we denote by $(dL,\, d \overset{\eta_d}{\to} (dL)R)$, where $dL \in |C|$. 

This correspondence provides us, for each arrow $d \overset{\varphi}\to cR$, with an arrow $dL \to c$, which we may denote by $^*\varphi$. And of course we can go back: given an arrow $dL \overset{f}{\to} c$, we can send $f$ to $\eta_d \,; (f)R$, and we may denote this arrow by $f^*$. I claim that right-asterisk and left-asterisk are inverses, and so we have a bijection between $C(dL, c)$ and $D(d, cR)$ for each $c \in |C|$ and $d \in |D|$.

It's pretty easy for us to say explicitly what it means to take $f \mapsto f^*$ (as long as we know the $\eta$ maps): we have $f^* = \eta_d \, ; (f)R$. In fact, we can say a little more: if $dL \overset{f}{\to} c \overset{g}{\to} c'$, then we can tell what $(f \,; g)^*$ should be:

The diagram makes clear the helpful formula $(f \, ; g)^* = f^* \, ; (g)R$.

Mapping $\varphi \mapsto {} ^*\varphi$ is a little harder for us to explicitly write down at this point; all we have assumed is that there must exist a unique such map. However, there is one thing we know about $^*\varphi$: from the diagrams above it is certainly clear that $\eta_d \, ; (^* \varphi)R = \varphi$.

Let us note that the correspondence $d \leftrightarrow dL$ is in fact a functor $C \leftarrow D$.

Lemma: There is a unique functor $C \overset{L}{\leftarrow} D$ such that $(d)L = dL$ and such that $1_D \overset{\eta}{\Rightarrow}LR$ is a natural transformation. Proof: We know what $L$ must be on objects, and drawing the commutative diagram tells us what it must be on arrows:

The initial property of $\eta_d$ tells us that there is a unique arrow $(\varphi)L$ such that the diagram commutes. I claim that $L$, defined this way, is functorial. QED It turns out that given the situation above, and given an element $c \in |C|$, we can find a coreflection of $c$ along $L$, i.e. a final object in $L \downarrow c$. This property of $L$ will end up clarifying somewhat what it means to map $\varphi \mapsto {} ^* \varphi$. Proposition: Given $c \in |C|$, there is a map $(c)RL \overset{\epsilon_c}{\to} c$ such that $(cR, cRL \overset{\epsilon_c}{\to} c)$ is a final object in $L \downarrow c$. Suppose that there were such a map $\epsilon_c$. That would mean that for any map $(d)L \overset{f}{\to} c$, there must exist a unique map $d \overset{\varphi}{\to} (c)R$ such that $f = (\varphi)L ; \epsilon_c$. 

Dualizing this equation, we get $$f^* = ((\varphi)L \,; \epsilon_c)^* = \eta_d \,; (\varphi)LR \,; (\epsilon_c)R$$ $$ = \varphi \, ; \eta_{cR}\,; (\epsilon_c)R = \varphi\, ; \epsilon_c^*.$$ Staring at this for a moment, a clever idea arises: choose $\epsilon_c:= {} ^*\text{Id}_{cR}$. Then, since we know left-asterisk and right-asterisk are inverse, we will have $f^* = \varphi$, so $f = {}^* \varphi$. This shows existence and uniqueness of $\varphi$ when we choose $\epsilon_c:= {}^* \text{Id}_{cR}$. It also has the happy consequence of giving us an explicit form for the left-asterisk operation: $$^*\varphi = (\varphi)L \, ; {}^* \text{Id}_{cL} = (\varphi)L \, ; \epsilon_c.$$ By precomposing $\varphi$ with a map $d' \overset{\psi}{\to} d$ and looking at the diagram, we get the result $^*(\psi \, ; \varphi) = (\psi)L \, ; {}^* \varphi$.

like where a groupoid is a category whose arrows satisfy an invertibility condition, but also a category is a groupoid equipped with extra arrows