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Let $P$ be a finite poset, and $X$ be a topological space stratified by $P$, in the sense that $X$ is equipped with a continuous map $X \to P$ in the Alexandroff topology (or equivalently with a collection of strata $S_p \subseteq X, p \in P$ such that $\cup_{p \leq q} S_p$ is closed for all $p \in P$). In Higher Algebra A.6.2, Lurie defined a stratified nerve $${\rm Sing}^P X \subseteq {\rm Sing} X$$ to be the subsimplicial set consisting of simplicies $f: \Delta^n \to X$ such that there exists a chain $p_0 \leq \dots \leq p_n \in P$ with $f(\Delta^{\{0,1, \dots, i\}} - \Delta^{\{0, \dots, i-1\}}) \subseteq S_{p_i}$.

If ${\rm Sing}^P X$ is an $\infty$-category, is the canonical map $|{\rm Sing}^P(X)| \to X$ a weak homotopy equivalence?

This is equivalent to asking whether ${\rm Sing}^P(X) \to {\rm Sing}(X)$ a weak homotopy equivalence of simplicial sets (in the Kan Quillen model structure), or more abstractly whether ${\rm Sing}(X)$ is the localization of ${\rm Sing}^P(X)$ at all maps.

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  • $\begingroup$ You probably already found this, but at least when $X$ is locally of singular shape, this is Corollary A.9.4 in Higher Algebra. Or are you specifically interested in generalisations beyond this case? $\endgroup$ Commented Oct 3, 2023 at 12:33
  • $\begingroup$ @ R. van Dobben de Bruyn Yes, I was interested in cases where we don't know that $X$ is of singular shape. $\endgroup$ Commented Nov 14, 2023 at 14:36

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This is false if the stratification is bad. (I first posted a more difficult example, but edited for simplification.)

Example. Let $X = \mathbf R$, let $Z$ be the closure of $\big\{\tfrac{1}{n}\ \big|\ n \in \mathbf Z_{>0}\big\}$, and let $U$ be its complement. Then $X$ has a stratification $\pi \colon X \to P := [1]$ with $X_0 = Z$ and $X_1 = U$. We know that $\operatorname{Sing}(X)$ is a contractible Kan complex.

On the other hand, let $Y = \mathbf R_{\leq 0} \amalg \mathbf R_{>0}$, with its natural map $f \colon Y \to X$. This induces a stratification $Y \to X \to [1]$ on $Y$, which is now a conical stratification (unlike $X \to [1]$). In particular, [HA, Thm. A.6.4(2) and Cor. A.9.4] show that $\operatorname{Sing}^P(Y)$ is an $\infty$-category and $\operatorname{Sing}^P(Y) \to \operatorname{Sing}(Y)$ is a weak homotopy equivalence.

We will show that the map $f_* \colon \operatorname{Sing}^P(Y) \to \operatorname{Sing}^P(X)$ induced by $f \colon Y \to X$ is an isomorphism of simplicial sets. Since $\operatorname{Sing}(Y)$ is the disjoint union of two contractible Kan complexes, this shows that $\operatorname{Sing}^P(X)$ is not weakly homotopy equivalent to $\operatorname{Sing}(X)$.

Since $f$ is a monomorphism (of $P$-stratified topological spaces), we get injectivity of $f_*$. For surjectivity, let $\sigma \colon \lvert\Delta^n\rvert \to X$ be an exit $n$-simplex, and let $j \in [1]$ be the highest stratum it meets, i.e. $j = \pi(\sigma(0,\ldots,0,1))$. By definition of $\operatorname{Sing}^P(X)$, there exists $i \in \{0,\ldots,n\}$ such that $\lvert \Delta^n \setminus \Delta^{\{0,\ldots,i-1\}} \rvert = \sigma^{-1}(X_j)$ (where we set $\Delta^{\{0,\ldots,i-1\}} = \varnothing$ if $i = 0$). Since $\lvert \Delta^n \setminus \Delta^{\{0,\ldots,n-1\}}\rvert$ is connected, it lands in a connected component $V$ of $X_j$, hence $\lvert \Delta^n \rvert$ lands in the closure $\overline V$ of $V$. But the closure $\bar V$ of any component of $X_0$ or $X_1$ is wholly contained in either $\mathbf R_{\leq 0}$ or $\mathbf R_{>0}$, so $\sigma$ lifts to an exit $n$-simplex $\tilde \sigma \colon \lvert \Delta^n \rvert \to Y$. $\square$

Remark. I don't know what happens if the space is bad but the stratification is still conical. I first tried $X$ to be a topologist's sine (which is not locally of singular shape), but when the stratification is the obvious one (with $X_0 = \{0\} \times [0,1]$), the map $\operatorname{Sing}^P(X) \to \operatorname{Sing}(X)$ still seems to be a weak equivalence.


References.

[HA] J. Lurie, Higher algebra. Book draft, version from 18 September 2017.

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    $\begingroup$ Thanks, this simpler example is very illustrative. $\endgroup$ Commented Nov 14, 2023 at 21:57
  • $\begingroup$ Yeah, I learned a lot from this too — somehow the key feature is that paths from $0$ to the right pass through too many strata, so cannot be obtained as finite zigzags of exit paths. $\endgroup$ Commented Nov 14, 2023 at 22:25

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