When do Kan extensions preserve limits/colimits?

Question

I'm guessing the answer to this question is well-known:

Suppose that $Y:C \to P$ and $F:C \to D$ are functors with $D$ cocomplete. Then one can define the point-wise Kan extension $\mathbf{Lan}_Y\left(F\right).$ Under what conditions does $\mathbf{Lan}_Y\left(F\right)$ preserve colimits? Notice that if $C=P$ and $Y=id_C,$ then $\mathbf{Lan}_Y\left(F\right)=F,$ so this is not true in general. Would $F$ preserving colimits imply this?

Dually, under what conditions does a right Kan extension preserve limits?

Thank you.

Answer 1

The pointwise left Kan extension of $F$ along $Y$ is a coend of functors $Lan_{Y}(F) = \int^{x}P(Yx,-).Fx$ where each functor $P(Yx,-).Fx$ is the composite of the representable $P(Yx,-):P \to Set$ and the copower functor $(-.Fx):Set \to D$. As a coend (colimit) of the $P(Yx,-).Fx$ the left Kan extension preserves any colimit by each of these functors.

Now the copower functor $(-.Fx)$ is left adjoint to the representable $D(Fx,-)$ and so preserves all colimits, so that $P(Yx,-).Fx$ preserves any colimit preserved by $P(Yx,-)$. Therefore $Lan_{Y}(F)$ preserves any colimit preserved by each representable $P(Yx,-):P \to Set$ for $x \in C$.

If $Y$ is the Yoneda embedding we have $P(Yx,-)=[C^{op},Set](Yx,-)=ev_{x}$ the evaluation functor at $x$ which preserves all colimits, so that left Kan extensions along Yoneda preserve all colimits.

Or if each $P(Yx,-)$ preserves filtered colimits then left Kan extensions along $Y$ preserve filtered colimits.

I think this is all well known but don't know a reference.

Answer 2

$F$ preserving colimits doesn't imply that $\text{Lan}_Y(F)$ preserves colimits, even if all the categories are cocomplete.

Consider, for example, the case $C = D$ and $F = 1_C$. Then the left Kan extension $\text{Lan}_Y(1_C)$ exists if and only if $Y$ has a right adjoint, and if it does exist, it is the right adjoint of $Y$. (This is Theorem X.7.2 of Categories for the Working Mathematician.) Of course, $1_C$ preserves colimits, but right adjoints usually don't.

(From your notation, I guess you're generalizing from the case where $P$ is the category of Presheaves on $C$ and $Y$ is the Yoneda embedding. In that case, as I bet you know, $\text{Lan}_Y(F) = - \otimes F$ not only preserves colimits but has a right adjoint.)

Answer 3

John Bourke's and Buschi Sergio's answers are correct, but I'd like to put some perspective on them.

I'll call my functors $f:C\to A$ and $g\colon C\to B$, and I want to understand when $\mathrm{LKan}_g f: B\to A$ is colimit preserving. I'll assume that $A$ and $B$ are cocomplete and $C$ is small, so all Kan extensions are pointwise Kan extensions, i.e., satisfying $$ (\mathrm{LKan}_g f)(b) = \mathrm{colim}_{c,g(c)\to b} f(c). $$ (The colimit is indexed over the comma category $C\times_B B_{/b}$, but I'll generally write colimits using a representative object of the indexing category.)

I'll show that, for fixed $g$, that $\mathrm{LKan}_gf$ is colimit preserving for every $f$ if and only if it is so for a "universal example" of a left Kan extension along $g$, which turns out to be exactly the "restricted Yoneda functor" of $g$.

First look at the special case of $B=\mathrm{PSh}(C)$, with $g=\rho\colon C\to \mathrm{PSh}(C)$ the Yoneda embedding. Then $\widehat{f}:=\mathrm{LKan}_\rho f \colon \mathrm{Psh}(C)\to A$ is colimit preserving, satisfies $\widehat{f}\rho =f$, and is in fact the essentially unique colimit preserving extension of $f$ to the presheaf category. In particular, if $f=\rho\colon C\to A=\mathrm{PSh}(C)$ is also the Yoneda functor, then $\widehat{\rho}=\mathrm{Id}\colon \mathrm{PSh}(C)\to \mathrm{PSh}(C)$.

The functor $\widehat{f}$ is colimit preserving because it is actually a left adjoint. Its right adjoint is the restricted Yoneda functor $R_f\colon A\to \mathrm{Psh}(C)$, defined by $$ R_f(a) = [f(-), a]_C $$ (I'm using $[x,y]_C$ for the hom-set.) To see this, I need to describe a natural bijection between $[\widehat{f}(X), a]_A$ and $[X,R_f(a)]_{\mathrm{Psh(C)}}$, where $X\in \mathrm{PSh}(C)$. When $X=\rho(c)$ is a representable presheaf this is easy, since $\widehat{f}(\rho(c))=f(c)$, while for general $X$ use the facts that $X$ is tautologically a colimit of representables: $X=\mathrm{colim}_{c,\rho(c)\to X}\rho(c)$, while $\widehat{f}=\mathrm{LKan}_\rho f$ is a pointwise Kan extension so $\widehat{f}(X) = \mathrm{colim}_{c,\rho(c)\to X} f(c)$, so we can reduce to the case of representable $X$.

Here is the key claim: the adjoint $R_f$ to the left Kan extension $\widehat{f}=\mathrm{LKan}_\rho f$ is also a left Kan extension.

Claim. We have $R_f=\mathrm{LKan}_f \rho \colon A\to \mathrm{PSh}(C)$.

Proof. Again we have a pointwise Kan extension, so we can compute $$ (\mathrm{LKan}_f \rho)(a) = \mathrm{colim}_{c,f(c)\to a} [-,c]_C. $$ Evaluating this at any object $x\in C$ gives $$ (\mathrm{LKan}_f \rho)(a)(x) = \mathrm{colim}_{c,f(c)\to a} [x,c]_C = \mathrm{colim}_{c,x\to c, f(c)\to a} *. $$ The last colimit is the set of components of the indexing category, and is easily seen to be [f(c),a]_C, as desired.

Reversing the roles of $f$ and $g$, we can now answer the original question in the case that $f=\rho\colon C\to \mathrm{Psh}(C)=A$, and for general $g\colon C\to B$.

Corollary. We have $\mathrm{LKan}_g \rho = R_g\colon B\to \mathrm{PSh}(C)$, so it preserves colimits if and only if $[g(c),-]_B \colon B\to \mathrm{Set}$ preserves colimits for all $c\in C$.

To answer the question in entirety, observe that $R_g$ is a kind of universal example of a left Kan extension along $g$.

Proposition. Given functors $f\colon C\to A$ and $g\colon C\to B$, we have $$ \mathrm{LKan}_g f = \widehat{f} (\mathrm{LKan}_g \rho) = \widehat{f} R_g \colon A\to B, $$ where $\widehat{f}\colon\mathrm{PSh}(C)\to A$ is the colimit preserving extenison of $A$.

Proof. Because $\widehat{f}$ preserves colimits, both left Kan extensions are computed pointwise using colimits over the same indexing category, and $\widehat{f}\rho=f$.

So now we have an answer.

Theorem. Given $g\colon C\to B$, we have that $\mathrm{LKan}_gf$ preserves colimits for any functor $f$ if and only if $R_g\colon \mathrm{PSh}(C)\to B$ preserves colimits.

Proof. If $R_g$ preserves colimits, then the previous proposition gives that $\mathrm{LKan}_gf$ does too. For the only if direction, remember that $\mathrm{LKan}_g \rho =R_g$.

Note that this criterion is a condition on $g$ only. I don't know of any criterion which is given by a condition on $f$.

Answer 4

Let $(a_i: A_i\to A)_{i\in I}$ with $I\in Cat$ a universal cocone in a category $\mathcal{A}$, and let $H: \mathcal{B}\to \mathcal{A}$.

We ask when:

for any $F: \mathcal{B}\to \mathcal{C}$ such that:

exist $L:=Lan_H F$ punctually (or at least it exist for the objects $A,\ A_i$ of the above diagram i.e. exists $L(A_i):=\varinjlim_{(B, b)\in H\downarrow A_i} F(B)$ and $L(A):=\varinjlim_{(B, b)\in H\downarrow A} F(B)$).

we have that $L(A)=\varinjlim_i L(A_i)$.

Consider the colimit category $\widehat{HA}:=\varinjlim_i H\downarrow A_i$, the functors $F\circ \pi_{A_i}:H\downarrow A_i\to \mathcal{B}\to \mathcal{C}$ induce a funtor $\hat{F}: \varinjlim_i H\downarrow A_i \to \mathcal{B}$ and is not hard verify that $\varinjlim_i L(A_i)=\varinjlim_i \varinjlim_{(B, b)\in H\downarrow A_i} F(B)= \varinjlim_{(B, b)\in \widehat{HA}} F(B)= \varinjlim \widehat{F}$.

Then the natural morphisms $\phi: \varinjlim_i L(A_i)\to L(A)$ is induced by the natural functor $\Phi: \widehat{HA}\to H\downarrow A $, then $\phi$ is a isomorphism (for any $F$ such that..) iff the functor $\Phi$ is final i.e. iff each morphism $H(B)\to A $ has a factorization on some $H(B')\to A_i\to A$ (through a morphism $B\to B'$) and two such factorization are connected in $H\downarrow A$.