2
$\begingroup$

Suppose that we have a functor $F : \boldsymbol{\Delta}^\bullet \to \mathsf{C}$ with domain the full subcategory of simplicial sets given by representable functors. For example, for each $\Delta^n = \hom(n,-)$ we can assign to it its baricentric subdivision $\mathsf{sd} \Delta^n \in \mathsf{sSet}$, or its geometric realization $|\Delta^n| \in \mathsf{Top}$.

By the Yoneda embedding, we have a fully faithful injective on objects functor $i: \Delta^{op} \hookrightarrow \boldsymbol{\Delta}$, hence $F$ can be thought of as a simplicial object

$$ F : \Delta^{op} \to \mathsf{C}. $$

On the other hand, if $X$ is any simplicial set, we know that it is a colimit of representables

$$ X = \mathsf{colim}_{\Delta^n \to X} \Delta^n. $$

If $\mathsf{C}$ is cocomplete, the definition

$$ \widetilde{F}X := \mathsf{colim}_{\Delta^n \to X} F\Delta^n, \tag{1}$$

makes sense and gives an extension of $F$ to a functor $\widetilde{F} : \mathsf{sSet} \to \mathsf{C}$.

In other terms, we are using that simplicial sets are the free cocompletion of $\Delta$, and so this is the universal cocontinuous extension of $F$.

If I am not mistaken, since $Fk = F\Delta^k$, using the cone leg arrows the maps

$$ Fk \to F\Delta^k \hookrightarrow \mathsf{colim}_{\Delta^n \to X} F\Delta^k= \widetilde{F}\Delta^n $$

gives a natural transformation $\eta : F\Rightarrow \widetilde{F}i$. So, assuming the former is correct, my question is:

Is $(\widetilde{F},\eta)$ a left Kan extension of $F$ along $i$?

I would also be interested in knowing what happens when we consider right Kan extensions, if these coincide and if not, what other interesting extension constructions can be made.

$\endgroup$
1
  • 1
    $\begingroup$ Yes, it is the left Kan extension. The same formula can be used to construct both. $\endgroup$
    – Zhen Lin
    Commented Aug 30, 2020 at 13:07

1 Answer 1

4
$\begingroup$

The fact that every functor $F$ like yours, with cocomplete codomain, admits a (essentially unique ) extension to $sSet$ amounts to the universal property of the free cocompletion, yes; and yes, the extension (has a right adjoint, called the $F$-nerve) and is a left Kan extension, along the Yoneda embedding $y : \Delta \to {\sf sSet}$.

There is plenty of places where this is proved, but I can't help from the usual self-promotion: Theorem 3.1.1 here.

As for right extensions, that's another story: the opposite of the category of presheaves on $\Delta^{op}$, i.e. the category $[\Delta, {\sf Set}]^{op}$, exhibits the universal property of the free completion of $\Delta$, and the contravariant Yoneda embedding $y^\sharp : \Delta^{op}\to [\Delta, {\sf Set}]$ yield a continuous extension for every $G$ with complete domain.

Usually, even assuming $\sf C$ bicomplete, it is not the case that $\text{Lan}_y F \cong \text{Ran}_{y^\sharp} F$.

$\endgroup$
2
  • $\begingroup$ Hi Fosco, I was aware of your book on (co)end calculus already. It is probably the least expected place in which I have found quotes of Borges and Cortázar. I hope to find some time to read it in the future. For now, I am definitely checking out Theorem 3.1.1. Many thanks for your answer! $\endgroup$
    – qualcuno
    Commented Aug 30, 2020 at 14:42
  • 1
    $\begingroup$ "Resultado parcial: te quiero, resultado total: te amo" if you're aware of the way in which Rayuela was written, I bet Cortázar took this sentence from a conversation he heard from a mathematician. ;-) $\endgroup$
    – fosco
    Commented Aug 30, 2020 at 14:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .