Suppose that you want to look at the left Kan extension of a functor $F : \mathcal{C} \to \mathcal{A}$ along a functor $K : \mathcal{C} \to \mathcal{B}$. It is widely known that if the colimit of the canonical diagram $$ \mathrm{colim}(K \downarrow d \to \mathcal{C} \overset{F}{\rightarrow} \mathcal{A})$$ exists in $\mathcal{A}$ for all object $d$ of $\mathcal{D}$, then the wanted left Kan extension exists and its value on $d$ is given by this colimit. However, to establish that result the previous colimit has to exist for all object $d$ of $\mathcal{D}$.

This raises the following question. Suppose that the left Kan extension of $F$ along $K$ exists, call it $L : \mathcal{D} \to \mathcal{A}$, and suppose that

$$ \mathrm{colim}(K \downarrow d \to \mathcal{C} \overset{F}{\rightarrow} \mathcal{A})$$

exists for a particular object $d$ of $\mathcal{D}$, but not necessarily for all objects of $\mathcal{D}$ (that's the point). Is it still true that $$L(d) \cong \mathrm{colim}(K \downarrow d \to \mathcal{C} \overset{F}{\rightarrow} \mathcal{A} )$$ ?


1 Answer 1


The answer is no (I think -- non-pointwise Kan extensions are a pain and I may have messed something up!). I wouldn't lose too much sleep over this, though -- in practice, you never know that some functor is a Kan extension without knowing it's a pointwise one.

The non-pointwise Kan extension I gave here works for a counterexample. See there for more details.

The Kan extension looks like this: (The best intuition I can offer is that it was actually constructed to have $F$ fully faithful, but the comparison 2-cell $\eta$ non-invertible, which forces it to be a non-pointwise Kan extension.)

$\require{AMScd}$ \begin{CD} \mathbf{B}\mathbb{N} @>G>> C\\ @VFVV \nearrow L \\ \mathbf{B}\mathbb{N}_+ \end{CD}

Here $\mathbf{B}\mathbb{N}$ is the monoid $\mathbb{N}$ regarded as a one-object category, $\mathbf{B}\mathbb{N}_+$ is the same with an initial object adjoined, $F$ is the natural inclusion, and $C$ looks like this:

$G\bullet \overset{(\eta_n)_{n \in \mathbb{N}}}{\overset{\to}{\to}} L\bullet \overset{L!}{\leftarrow} L\emptyset$

Here $G\bullet$ is the fully faithful image of $G$ (so it's a copy of $\mathbf{B}\mathbb{N}$), and $L\emptyset \overset{L!}{\to} L\bullet$ is the fully faithful image of $L$ (so it's a copy of $\mathbf{B}\mathbb{N}_+$ where $L\emptyset$ is the initial object). $C(G\bullet, L\bullet)$ is generated by an arrow $\eta = \eta_0$ under the equation $\eta_n = \eta \circ Gn = Ln \circ \eta$. So $\eta$ constitutes a comparison natural transformation $G \implies LF$.

Why it's a left Kan extension:

An exhaustive analysis of the functors $H: \mathbf{B}\mathbb{N}_+ \to C$ (there are 5 families of them, depending on where the objects are sent) reveals that the only one which admits a natural transformation $G \implies HF$ or $L \implies H$ is $L$ itself, and this diagram is in fact a left Kan extension $L = \mathrm{Lan}_F G$.

Why it's a counterexample:

You can check that $Hom(G\bullet,-)$ and $Nat(Hom(F,\bullet),Hom(G,-))$ are isomorphic; they map $L\emptyset \mapsto \emptyset$, $L\bullet \mapsto \mathbb{N}$, $G\bullet \mapsto \mathbb{N}$. So $G\bullet$ "should" be the value of $\mathrm{Lan}_F G(\bullet)$. even though the value in the actual left Kan extension $L$ is $L\bullet$.

  • $\begingroup$ I am not sure that I understand what you mean by $\mathrm{Nat}(\mathrm{Hom}(F,\bullet),\mathrm{Hom}(G,-))$. Where do $\mathrm{Hom}(F,\bullet)$ and $\mathrm{Hom}(G,-)$ live? $\endgroup$
    – L.Guetta
    Commented Jul 19, 2017 at 13:25
  • $\begingroup$ I agree, my notation is confusing! $\bullet$ is the noninitial object of $\mathbf{B}\mathbb{N}_+$ here (although I'm also using it to denote the corresponding object of $\mathbf{B}\mathbb{N}$ elsewhere!). So $Hom(F,\bullet)$ is a functor $\mathbf{B}\mathbb{N}^\mathrm{op} \to \mathsf{Set}$. Likewise, for $c \in C$, $Hom(G,c)$ is a functor $\mathbf{B}\mathbb{N}^\mathrm{op} \to \mathsf{Set}$. Then $Nat(Hom(F,\bullet),Hom(G,-)$ is the functor $C \to \mathsf{Set}$ sending $c$ to the set of natural transformations from $Hom(F,\bullet)$ to $Hom(G,c)$.... $\endgroup$
    – Tim Campion
    Commented Jul 19, 2017 at 13:43
  • $\begingroup$ The point of this is that if $c_\ast = colim(F \downarrow \bullet \to \mathbf{B}\mathbb{N} \to C)$ (i.e. $c_\ast$ is what $Lan_F G(\bullet)$ is "supposed to be" by the pointwise formula), then $c_\ast$ is characterized by the fact that $Hom(c_\ast,-)$ is isomorphic to $Nat(Hom(F,\bullet),Hom(G,-))$. (I tend to think of this in terms of the colimit formula encoding a weighted colimit, although there are other ways to think about it.) So the computation is that $c_\ast = G \bullet$, but $Lan_F G(\bullet) = L \bullet$; the pointwise formula and non-pointwise Kan extension disagree. $\endgroup$
    – Tim Campion
    Commented Jul 19, 2017 at 13:52

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