Is the singular simplicial functor full

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.

• 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). – Denis Nardin Feb 11 '17 at 4:04
• @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(-)|). – Stefano Nicotra May 20 at 9:40
• @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 $|\mathrm{Sing}X|→X$... – Denis Nardin May 20 at 9:43

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 $X \to |X|$.