4
$\begingroup$

A functor $F:\mathcal{C}\to \mathcal{D}$, from an essentially small category to a cocomplete category induces a realisation-nerve adjunction between the categories $\mathbf{Fun}(\mathcal{C}^{op},\mathbf{Set})$ and $\mathcal{D}$. I vaguely recall reading that "the nerve functor deserves the name singular functor only if it is fully faithful". For the standard cosimplicial topological space $\Delta^\bullet_{top}:\mathbf{\Delta}\to \mathbf{Top}$, one retrieves the geometric realisation-singular simplicial adjunction. It is easy to see that singular simplicial functor is faithful. But, I do not see why it is full, if it is.

The singular simplicial functor is fully faithful iff the left Kan extension $\mathbf{Lan}_{\Delta^\bullet_{top}}\Delta^\bullet_{top}$ is point-wise and is isomorphic to the identity functor. Hence, I am 'almost-convinced' that the singular simplicial functor is not full, and an alternative question would be:

what is an example of a topological space $X$, for which $\mathbf{Lan}_{\Delta^\bullet_{top}}\Delta^\bullet_{top}(X)\ncong X$?

PS I assumed earlier that my question should be well-known, and I have asked this question on math.stackexchange.com, but it does not seem to be the case.

Improve this question
$\endgroup$
4
  • $\begingroup$ I suspect that the left Kan extension is the identity for numerically generated spaces (which are the kind of spaces you want to consider in homotopy theory anyway). $\endgroup$
    – Denis Nardin
    Commented Feb 11, 2017 at 4:04
  • $\begingroup$ @DenisNardin Is that true? I thought that every numerically generated space was a canonical colimit over ∆, where we consider all possible morphisms between standard simplices and not just the realisations of the ones coming from ∆. Besides, unless I'm missing something, the Kan extension described in the Question should be just |Sing(X)|, which is a CW-complex so any topological space which is not a CW-complex should be a counterexample (and even if you start with a CW-complex you usually end up with a much bigger one after applying |Sing(-)|). $\endgroup$
    – Stefano Nicotra
    Commented May 20, 2019 at 9:40
  • $\begingroup$ @StefanoNicotra It might be false, you're right that looking at it today it seems a lot fishier to me (maybe I was thinking of homotopical Kan extensions when I wrote that comment...). In particular the counit is almost certainly $|\mathrm{Sing}X|→X$... $\endgroup$
    – Denis Nardin
    Commented May 20, 2019 at 9:43
  • $\begingroup$ Is there a positive result that under some assumptions $\text{Sing} X$ is full? e.g. that for CW complexes $X$ and $Y$ $Hom_{Top}(X,Y)=Hom_{sSets}(Sing X, Sing Y)$? $\endgroup$
    – user494312
    Commented Feb 21, 2023 at 14:17

1 Answer 1

Reset to default
6
$\begingroup$

Any non-discrete totally disconnected space gives a counterexample, e.g. the Cantor space $2^{\mathbb{N}}$.

If a space $X$ is totally disconnected, then every simplex in it is constant (since the simplices are connected). So the singular set of $X$ is the discrete simplicial set on its points $|X|$; hence its geometric realisation is the discrete space on its points.

Converting this explicitly to a counterexample to fullness: consider the identity map $|X| \to X$, where $X$ is totally disconnected and non-discrete. Then the induced map of singular sets is an isomorphism; but its inverse doesn’t come from any continuous map $X \to |X|$.

Improve this answer
$\endgroup$

You must log in to answer this question.

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