Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
For the existence of $\mathrm{Lan}_Y(F)$, we should require that $C$ is essentially small. –  Martin Brandenburg Sep 22 '13 at 12:16
add comment

3 Answers 3

up vote 13 down vote accepted

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 ref.

share|improve this answer
    
Thanks, this is quite helpful! –  David Carchedi Nov 12 '12 at 21:21
add comment

$F$ 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.)

share|improve this answer
    
Thanks Tom. Do you know under what general conditions the left Kan extension does preserve colimits? Of course I am aware that this holds when $P$ is the presheaf category, but I would like to know this more generally. Also, what is known about right Kan extensions along the Yoneda embedding (into a complete category)? –  David Carchedi Nov 12 '12 at 8:32
    
(right Kan extensions along Yoneda, on a category with finite limits, such that the functor I am extending also preserves finite limits) –  David Carchedi Nov 12 '12 at 13:52
add comment

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 $H\downarrow A$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.