Left Adjoint of the Nerve Functor – Higher Category Theory

Question

I recently stumbled upon a formula for the left adjoint of the nerve functor. Let $X$ and $Y$ be simplicial sets, then:
\begin{equation}
\mathbf{sSet}(X,Y)
\cong\mathbf{sSet}(\varinjlim_{\Delta^n\rightarrow X}\Delta^n,Y)
\cong\varprojlim_{\Delta^n\rightarrow X}\mathbf{sSet}(\Delta^n,Y)
\cong\varprojlim_{\Delta^n\rightarrow X}Y_n.
\end{equation}
Let $\mathcal{C}$ be a category and view $[n]$ as a category induced by $\leq$, then:
\begin{equation}
\mathbf{sSet}(X,N\mathcal{C})
\cong\varprojlim_{\Delta^n\rightarrow X}N_n\mathcal{C}
\cong\varprojlim_{\Delta^n\rightarrow X}\mathbf{Cat}([n],\mathcal{C})
\cong\mathbf{Cat}(\varinjlim_{\Delta^n\rightarrow X}[n],\mathcal{C}).
\end{equation}
Therefore the left adjoint of the nerve is given by:
\begin{equation}
\tau(X)=\varinjlim_{\Delta^n\rightarrow X}[n].
\end{equation}
I'm not sure, if this formula is correct though. I did not find it anywhere in "Higher Categories and Homotopical Algebra" (See here, Nerves are introduced in section 1.4. on page 14.) by Deniz-Charles Cisinski or somewhere else yet. Is it correct? If no, where is my error? If yes, where else can I find it? To continue the calculation, since $N_n\mathcal{C}=\mathbf{sSet}(\Delta^n,N\mathcal{C})\cong\mathbf{Cat}([n],\mathcal{C})$, we have:
\begin{equation}
\tau N\mathcal{C}
=\varinjlim_{\Delta^n\rightarrow N\mathcal{C}}[n]
\cong\varinjlim_{[n]\rightarrow\mathcal{C}}[n]
\cong\mathcal{C}.
\end{equation}
A direct conclusion of this using the adjunction would be the nerve being fully faithful in agreement with HCaHA, Proposition 1.4.11. on page 17. I'm not sure, if the last isomorphism is correct though. An analogy fails. Let $Z$ be a topological space, then $\operatorname{Sing}_n(Z)=\mathbf{sSet}(\Delta^n,\operatorname{Sing}(Z))\cong\mathbf{Top}(|\Delta^n|,Z)$, but:
\begin{equation}
|\operatorname{Sing}(Z)|
=\varinjlim_{\Delta^n\rightarrow\operatorname{Sing}(Z)}|\Delta^n|
\cong\varinjlim_{|\Delta^n|\rightarrow Z}|\Delta^n|
\cong Z
\end{equation}
obviously isn't true in general. Is the last isomorphism in the first equation correct? If yes, why does it fail in the second? What are the conditions for it to hold?

Answer 1

Before anything else, let me point out that you have some typos with some $\varinjlim$'s that should be $\varprojlim$'s.

Otherwise, the first and second formula are correct, but hardly usable in practice. In fact, you typically deduce the second one from a more usable version of the first one - there are probably other proofs, but to understand why it's not that easy, note that colimits of categories are very badly behaved/understood.

As to "where you can find it", note that it is a consequence of the following three facts : 1- $\tau$ is a left adjoint, so preserves colimits; 2- $X\cong \varinjlim_{\Delta^n\to X}\Delta^n$; 3- $\tau(\Delta^n) = [n]$.

1- and 3- are relatively straightforward; and 2- is a classical consequence of the Yoneda lemma (in fact you use it in your first equation $\mathbf{sSet}(X,NC)\cong \varprojlim_{\Delta^n\to X}N_nC$).

So I don't think it will be stated in this case often because it is a very general fact (little to do with $N$ specifically) - so it will be stated in more generality- and hardly useful for this specific adjunction because there is a more concrete description of $\tau$.

For your second question, the nerve being fully faithful is essentially equivalent to the canonical map $\varinjlim_{[n]\to C}[n]\to C$ being an equivalent, so I'm not sure how to deduce it this way - again, colimits of categories are complicated. You can prove directly $\tau N C\cong C$ using the description of $\tau X$ in terms of the "homotopy category of $X$".

You are correct that this is not the case in general for topological spaces (for example, if $X$ is totally disconnected, then $X^{discrete}\to X$ induces an equivalence $\{|\Delta^n|\to X^{discrete}\}\to \{|\Delta^n|\to X\}$ so this cannot distinguish totally disconnected from discrete.

I don't know if you can state general conditions for it to hold that aren't just "it does". I mean, there is an interesting one is that is that it is equivalent to the "nerve" functor being fully faithful; but you already knew that.