Extending an adjunction using colimits

Question

I'd like a confirmation of a fact about colimits and adjunction; the motivation is that I think that this fact is used implicitly sometimes in Kerodon, but I've never seen it written out. $\require{AMScd}$

Let $C'\subset C$ and $D$ be categories; assume that $D$ is cocomplete and that the inclusion is full and preserves colimits. Let $G':D\to C'$ be a functor, and let $G:D\to C$ be the functor $G'$ followed by the inclusion. Finally, let $F':C'\to D$ be a functor left adjoint to $G'$. Observations:

1. If, fixed any $c\in C$, there is an isomorphism $D(Fc,-)\to C(c,G-)$ in $Set^D$, for some $Fc\in D$, the rule $c\mapsto Fc$ extends to a unique functor $F:C\to D$ left adjoint to $G$.
2. Given $f:c_2\to c_1$ in $C$, the morphism $Ff$ obtained in point 1 is the unique that makes the following square commutative: $$\begin{CD}D(Fc_1,-)@>\sim>>C(c_1,G-)\\ @VV-\circ FfV @VV-\circ fV\\ D(Fc_2,-)@>\sim>>C(c_2,G-) \end{CD}$$
3. Fixed any $c'\in C'$, clearly $C'(c',G'-)= C(c',G-)$ in $Set^D$.
4. The functor $(-\circ G):Set^{C}\to Set^D$ preserves limits.

Claim. Assume that every $c\in C$ is equal to $\operatorname{colim}_{i\in I} c_i$, for some small category $I$ which depends on $c$, indexing a diagram in $C'\subset C$ (so the $c_i$ are in $C'$). Then $F'$ extends to a functor $F:C\to D$ left adjoint to $G$.

Proof. Fix $c\in C$, so that $c=\operatorname{colim}_{i\in I} c_i$, and all $c_i$ are in $C'$. For any $c_i$ there is an isomorphism $\gamma_i:D(F'c_i,-)\to C'(c_i,G'-)=C(c_i,G-)$ in $Set^D$. These $\gamma_i$ form a morphism of diagrams in $Set^D$: in fact, if $h:i\to j$ is an arrow in $I$ and $f_h:c_j\to c_i$ is its image in $C'\subset C$, the following squares commute thanks to the adjunction $F'\dashv G'$:$$\begin{CD}D(F'c_i,-)@>\gamma_i>>C'(c_i,G'-)@=C(c_i,G-)\\ @VV-\circ Ff_hV @VV-\circ f_hV @VV-\circ f_hV\\ D(F'c_j,-)@>\gamma_j>>C'(c_j,G'-)@=C(c_j,G-) \end{CD}$$

Since the operation of limit is functorial, there is an isomorphism $\lim_i D(F'c_i,-)\to \lim_i C(c_i,G-)$ in $Set^D$.

Surely there is an isomorphism $\lim_i D(F'c_i,-)\to D(\operatorname{colim}_iF'c_i,-)$ in $Set^D$, which commutes with the canonical morphisms. There is as well an isomorphism $\lambda:\lim C(c_i,-)\to C(\operatorname{colim} c_i,-)=C(c,-)$, such that this canonical triangle in $Set^{C}$ commutes for all $i$:$$\begin{CD} \lim C(c_i,-)@>\lambda>>C(c,-)\\ @VVV @VVV\\ C(c_i,-) @= C(c_i,-) \end{CD}$$ Therefore the same triangle precomposed by $G$ commutes in $Set^D$:$$\begin{CD} \lim C(c_i,G-)@>\lambda G>>C(c,G-)\\ @VVV @VVV\\ C(c_i,G-) @= C(c_i,G-) \end{CD}$$ By point 4 the left column is indeed made by the limit of the diagram $(C(c_i,G-))_i$ with its canonical morphism, so $\lambda G$ is an isomorphism commuting with the canonical morphisms.

At this point, choose once and for all a small category $I$ for each $c\in C$, such that $c=\operatorname{colim}_{i\in I} c_i$, with all the $c_i$ in $C'$; in particular for those $c\in C'$ choose $I$ as the terminal category, i.e. express $c$ as a trivial colimit of itself. By the previous arguments, setting $Fc:=\operatorname{colim}_{i\in I} F'c_i$ on the objects $c\in C$, there is an isomorphism $D(Fc,-)\to C(c,G-)$ in $Set^D$ for all $c$.

By point 1, the only thing left to check is that the definition that we will get for $F$ on morphisms agrees with $F'$. So let $f':c'_2\to c'_1\in C'$. The "square" $$\begin{CD} D(Fc'_1,-)@>\sim>>C(c'_1,G-)\\ @. @VV-\circ f'V\\ D(Fc'_2,-)@>\sim>>C(c'_2,G-) \end{CD}$$ by our definitions is exactly the "square" $$\begin{CD} D(F'c'_1,-)@>\sim>>C'(c'_1,G'-)\\ @. @VV-\circ fV\\ D(F'c'_2,-)@>\sim>>C'(c'_2,G'-) \end{CD}$$ so the morphism $Fc'_2=F'c'_2\to F'c'_1=Fc'_1$ that we get is $F'f'$ by point 2.

Regarding Kerodon: look for example at Proposition 1.1.6.19. It really seems, to me, that the proof suggested by the author is to apply the claim above with: $C'\subset C$ the inclusion of connected simplicial sets into simplicial sets, $D:=Set$, $G$ the functor sending a set $J$ to the constant simplicial set over $J$, and $F'$ the singleton constant functor.

Answer 1

Assume $F\dashv G$, and $i : C\to C'$ is fully faithful(1) and dense(2).

This is slightly more than what you ask: every object of $C'$ is a colimit of objects of $C$ in a way that you can control. Now, assume that the extension $F' : C'\to D$ is the left extension of $F$ along $i$ (so again, this is slightly more: $F'$ is not an extension, it is a canonical extension).

Because of (1), $Lan_iF\circ i\cong F$, and because of (2), $Lan_ii=1_{C'}$. But then, you can obtain a candidate unit and counit for $Lan_iF\dashv iG$ as follows.

1. Counit: compose $$ Lan_i F \circ i \circ G \cong F\circ G \overset\epsilon\Rightarrow 1$$

2. Unit: compose $$ i\overset{i*\eta}\Rightarrow i\circ G\circ F \cong i\circ G\circ Lan_i F\circ i$$ which mates to $\bar \eta : Lan_ii\Rightarrow i\circ G\circ Lan_i F$, i.e. to $1_{C'}\cong Lan_ii\overset{\bar\eta}\Rightarrow i\circ G\circ Lan_i F $ thanks to (2).

I am confident the zig-zag identities for $Lan_iF\dashv iG$ can be proved with some patience...

Answer 2

$\newcommand{\D}{\mathfrak{D}}\newcommand{\C}{\mathsf{C}}\newcommand{\colim}{\operatorname{colim}}$Fix a left-right adjunction $F:\C\rightleftarrows \D:G$ where $\C$ is a full subcategory of $\C'$. Let $\iota:\C\hookrightarrow\C'$ be the inclusion functor. We assume every object $\varsigma\in\C'$ has an associated diagram $\Sigma(\varsigma):J(\varsigma)\to\C$ with $\varsigma\cong\colim(\iota\Sigma(\varsigma))$, for some (small) category $J(\varsigma)$, and we define $G':=\iota\circ G$. Assume for sake of logical coherency that we have fixed the data of the pairings $(\varsigma,\Sigma(\varsigma))$, perhaps $\Sigma$ and $J$ are thought to be functions of some kind. Note if $\varsigma\in\C$ we can choose $\Sigma(\varsigma)$ to be the constant diagram to $\varsigma$ from the singleton category. We also need to assume $\D$ is cocomplete, or at least that the colimits we refer to in $\D$ exist.

N.B. the author has since reversed the roles of $\C',\C,F',F$ in their post. My notation differs from theirs accordingly, but this is an answer to the same problem.

---

You want to define a left adjoint $F'$ to $G'$ which extends $F$. I would define $F'(\varsigma):=\colim(F\Sigma(\varsigma))$ on objects. We only need to define an initial morphism $\mu_{\varsigma}:\varsigma\to GF'(\varsigma)=G'F'(\varsigma)$ for all $\varsigma\in\C'$: from this, $F'$ would be uniquely extended to a functor in such a way that $F'\dashv G'$.

The arrows (ranging over $j\in J(\varsigma)$) $\Sigma(\varsigma)(j)\overset{\eta}{\longrightarrow}GF(\Sigma(\varsigma)(j))\to GF'(\varsigma)$, where $\eta$ is the unit of $F\dashv G$, assemble to a cocone under $\Sigma(\varsigma)$ and so a unique arrow $\mu_{\varsigma}$ is defined. Is it universal? Suppose $\partial\in\D$, $\varsigma\in\C'$ and an arrow $f:\varsigma\to G'(\partial)=G(\partial)$ is given. We want to show there is a unique $g:F'(\varsigma)\to\partial$ with $G(g)\mu_{\varsigma}=f$.

Any such $g$ would be uniquely given by some cocone $(g_j:F\Sigma(\varsigma)(j)\to\partial)_{j\in J(\varsigma)}$ and $G(g)\mu_{\varsigma}$ can be seen to equal the arrow uniquely given by the cocone $(G(g_j)\eta_{\Sigma(\varsigma)(j)}:\Sigma(\varsigma)(j)\to G(\partial))_{j\in J(\varsigma)}$. We then want to say, there is a unique family of arrows $g_j$ for which that cocone corresponds to $f$.

But under $F\dashv G$, it is clear that the $g_j$ are the transposes of $\Sigma(\varsigma)(j)\to\varsigma\overset{f}{\longrightarrow}G(\partial)$ and are thus uniquely determined and exist (that $\C$ is a full subcategory ensures the composite is a genuine arrow of $\C$, so to take its transpose under $F\dashv G$ makes sense). Transposition preserves the cocone property, so we are ok on that front.

That (uniquely) creates an adjunction $F'\dashv G'$ with unit $\mu$, as desired. We can unravel the definitions slightly to figure out whether or not $F'$ really extends $F$:

On objects of $\C$, we chose $\Sigma$ to be the constant diagram to that object, so $F'\circ\iota=F$ on objects.

$F'(f)$ for general arrows $f:\varsigma\to\varsigma'$ is defined as the unique arrow induced by $\varsigma\overset{f}{\longrightarrow}\varsigma'\overset{\mu_{\varsigma'}}{\longrightarrow}G'F'(\varsigma')$. If $f$ is an arrow in $\C$, then all diagrams involved are trivial and this unique arrow is exactly the transpose of $\eta_{\varsigma'}f$, which is just $F(f)$, so $F'\circ\iota=F$ on arrows too.

Answer 3

As Fosco has suggested, this is true if we assume that the colimits are suitably coherent and a certain pointwise extension exists, which is true if $D$ admits enough colimits. I shall give an easy proof using the theory of relative adjunctions.

Proposition. Let $F : C \rightleftarrows D : G$ be an adjunction and let $I \colon C \hookrightarrow C'$ be a dense and fully faithful functor. Suppose that the pointwise left extension $F' \colon C' \to D$ of $F$ along $I$ exists. Then $F' \dashv I G$.

Proof. Since $F \dashv G$ and $I$ is fully faithful, we have an $I$-relative adjunction $F {}_I \dashv I G$. Since $I$ is dense, we therefore have $F' \dashv IG$ by Proposition 5.4.7 of Arkor's Monadic and higher-order structure.

Answer 4

What you want follows easily from an appropriate understanding of the preservation of colimits by left adjoints.

---

Consider a functor $G\colon D\to C$ and a diagram $\Sigma\colon E\to C$ for which each object $e$ of $E$ has a chosen reflector of $\Sigma(e)$ across $G$, i.e. a morphism $\eta_e\colon\Sigma(e)\to GFe$ such that any morphism $\Sigma(e)\to Gd$ factors as $Gf\circ\eta_e\colon\Sigma(e)\to GFe\to Gd$ for a unique morphism $f\colon Fe\to d$.

Note that we can uniquely extend the assignment of objects $e\mapsto Fe$ to a diagram $F\colon E\to C$ so that $\eta\colon\Sigma\to GF$ is a natural transformation. Indeed, for a morphism $e\to e'$ in $E$, we have that $\Sigma(e)\to\Sigma(e')\to GFe'$ must factor as $\Sigma(e)\to GFe\to GFe'$ for a unique morphism $Fe\to Fe'$.

Moreover, it is straightforward to check that the above extension makes $F$ into a functor. In particular, if $\Sigma$ is the identity functor $C\to C$, we obtain the left adjoint $F\colon C\to D$ to $G\colon D\to C$.

Consider now a cocone with vertex $Gd$ under the diagram $\Sigma$. Each component $\Sigma(e)\to Gd$ factors as $\Sigma(e)\to GFe\to Gd$ for a unique morphism $Fe\to d$. Moreover, the morphisms $Fe\to d$ form a cocone with vertex $d$ under the diagram $F\colon E\to C$. Indeed, $\Sigma(e)\to Gd$ factors as $\Sigma(e)\to\Sigma(e')\to Gd$, and hence as $\Sigma(e)\to\Sigma(e')\to GFe'\to Gd$, which by naturality is the same as $\Sigma(e)\to GFe\to GFe'\to Gd$. Thus $Fe\to Fe'\to d$ equals $Fe\to d$.

Thus we have a bijective correspondence between cocones under $\Sigma\colon J\to C$ with verted $Gd$ and cocones under $F\colon J\to D$ with vertex $d$. Moreover, this correspondence is natural in $d$, i.e. post-compositing with $d\to d'$ corresponds to post-composing with $Gd\to Gd'$.

Finally, if $c=\operatorname{colim}_e\Sigma(e)$, then morphisms $c\to Gd$ correspond naturally to cocones under $F$ with vertex $d$. It follows that a morphism $c\to Gd$ is a reflector of $c$ across $G\colon D\to C$ if and only if the corresponding cocone with components $Fe\to d$ under $F\colon E\to D$ is a universal cocone.

It follows that a necessary and sufficient condition for a left adjoint to extend along a dense fully faithful functor is that, for each object to which the left adjoint is to be extended, there is a diagram realizing the object as a colimit, and for which the corresponding diagram of values of the left adjoint functor also has a colimit.

---

One can generalize the above argument to a $2$-categorical setting by replacing colimits of diagrams with left Kan extensions, and reflectors of each object in the value of the diagram with relative left adjoints. Something I've written before in that direction is https://math.stackexchange.com/a/1867749/400.

Answer 5

After thinking about this for a while, it seems to me that the right adjoint plays an unexpectedly large role in ensuring that the extension is well defined. Let me state a precise theorem for the record:

Theorem. Let $i : \mathcal{C}' \to \mathcal{C}$ be a fully faithful functor and let $F' : \mathcal{C}' \to \mathcal{D}$ be a functor. Assume the following hypotheses:

• $F'$ has a right adjoint, say $G' : \mathcal{D} \to \mathcal{C}'$.

• There is an ordinal-indexed sequence $$\mathcal{C}_0 \subseteq \mathcal{C}_1 \subseteq \mathcal{C}_2 \subseteq \cdots \subseteq \mathcal{C}$$ of full subcategories such that:

  • $\mathcal{C}_0$ is the image of $i$.

  • For every ordinal $\alpha$ and every object $c$ in $\mathcal{C}_{\alpha + 1}$, there exist a diagram $c' : \mathcal{J} \to \mathcal{C}_\alpha$ and a colimit cocone $c' \Rightarrow c$ in $\mathcal{C}$, and $\mathcal{D}$ has colimits of diagrams of shape $\mathcal{J}$.

  • For every limit ordinal $\lambda$, $\mathcal{C}_\lambda = \bigcup_{\alpha < \lambda} \mathcal{C}_\alpha$.

  • $\mathcal{C} = \bigcup_\alpha \mathcal{C}_\alpha$.

Then:

• $i G' : \mathcal{D} \to \mathcal{C}$ has a left adjoint, say $F : \mathcal{C} \to \mathcal{D}$.

• We have an isomorphism $F' \cong F i$.

The hypotheses can be improved, but it gets complicated to state. The proof is basically as you say. But if, instead of assuming $F'$ has a right adjoint and demanding $F$ have "the same" right adjoint, we only demand $F$ preserve (some) colimits and have an isomorphism $F' \cong F i$, things can go wrong. For example, take $\mathcal{C} = \textbf{Ab}$, $\mathcal{D} = \textbf{Set}$, and let $\mathcal{C}'$ be the full subcategory containing just $\mathbb{Z}$. Then we can take $\mathcal{C}_1$ to be the full subcategory of free abelian groups and $\mathcal{C}_2 = \mathcal{C}_3 = \cdots = \textbf{Ab}$. Consider the functor $F' : \mathcal{C}' \to \mathcal{D}$ sending $\mathbb{Z}$ to $1$. There is no colimit-preserving functor $F : \mathcal{C} \to \mathcal{D}$ extending $F'$: any functor $\textbf{Ab} \to \textbf{Set}$ that preserves initial objects must send everything to $\emptyset$ too.

Perhaps the way to understand this phenomenon is to think profunctorially. Given $F' : \mathcal{C}' \to \mathcal{D}$ and $i : \mathcal{C}' \to \mathcal{C}$, we would seek a functor $\tilde{F} : \mathcal{C}^\textrm{op} \times \mathcal{D} \to \textbf{Set}$ such that:

• For all $c'$ in $\mathcal{C}'$ and $d$ in $\mathcal{D}$, $\tilde{F} (i (c'), d) \cong \mathcal{D} (F' (c'), d)$ naturally.

• For all $c$ in $\mathcal{C}$, $\tilde{F} (c, -) : \mathcal{D} \to \textbf{Set}$ is representable.

• For all $d$ in $\mathcal{D}$, $\tilde{F} (-, d) : \mathcal{C}^\textrm{op} \to \textbf{Set}$ preserves limits (i.e. sends colimits in $\mathcal{C}$ to limits in $\mathcal{D}$).

Given such $\tilde{F}$, there is a unique (up to unique isomorphism) functor $F : \mathcal{C} \to \mathcal{D}$ such that $\tilde{F} (c, d) \cong \mathcal{D} (F (c), d)$ naturally, and $F$ will preserve colimits. In the case where $F'$ has a left adjoint $G' : \mathcal{D} \to \mathcal{C}'$ and $i : \mathcal{C}' \to \mathcal{C}$ is fully faithful, we have $$\mathcal{D} (F' (c'), d) \cong \mathcal{C}' (c', G' (d)) \cong \mathcal{C} (i (c'), i (G' (d)))$$ so $\tilde{F} (c, d) = \mathcal{C} (c, i (G' (d)))$ satisfies the requirements except possibly for representability of $\tilde{F} (c, -)$. Without the right adjoint $G'$ and the fully faithful $i$, we could not make such a definition and would instead have to consider something like, say, $$(\operatorname{Lan}_i F')^\sim (c, d) = \int_{c' : \mathcal{C}'} \textrm{Hom} (\mathcal{C} (i (c'), c), \mathcal{D} (F' (c'), d))$$ which may or may not coincide with $\mathcal{C} (c, i (G' (d)))$. The canonical map $\mathcal{D} (F' (c'), d) \to (\operatorname{Lan}_i F')^\sim (i (c'), d)$ is not guaranteed to be a bijection either. Of course, as the notation suggests, if $(\operatorname{Lan}_i F')^\sim (c, -) : \mathcal{D} \to \textbf{Set}$ is representable for every $c$ then we have a (pointwise!) left Kan extension $\operatorname{Lan}_i F' : \mathcal{C} \to \mathcal{D}$ of $F'$ along $i$. Again, this may or may not coincide with $F$. (They coincide if $i$ is a dense functor, which means that the canonical map $$\mathcal{C} (c_0, c_1) \to \int_{c' : \mathcal{C}'} \textrm{Hom} (\mathcal{C} (i (c'), c_0), \mathcal{C} (i (c'), c_1))$$ is bijection. Density is a somewhat stronger condition than generation under colimits.)