A couple of days back I asked a question Is there a Geometric/Smooth version of Homotopy Hypothesis using the path $\infty$-Groupoid of a Smooth Space? in **MO** about the existence of a possible Smooth/Geometric version of *Homotopy Hypothesis* using the notion of Path $\infty$-groupoid of a smooth space.

*After a discussion in the comments section with @David Roberts I got a feeling (but not completely convinced) that though Path 1-groupoid and smooth fundamental 1-groupoid of a smooth space are quite different objects but "if we move upto infinity level" and present them as Kan Complexes then they are becoming the same object.*

3 months back I asked the following **MO** question What is the geometric realization of the the nerve of a fundamental groupoid of a space?.

From the discussions in

**now I have the following Questions/Doubts:**

We know that the construction of Smooth Fundamental 1-Groupoid and Path 1-Groupoid of a smooth space induce natural functors $Man \rightarrow Groupoids$. Now from the discussion in What is the geometric realization of the the nerve of a fundamental groupoid of a space? I expect that $|N \circ \pi_{\leq 1}(X)|$ contains all informations of the 1st Homotopy groups of the smooth space $X$ where $N$ is the *Nerve* functor, $\pi_{\leq 1}$ is the *Smooth Fundamental 1-Groupoid functor* and $|-|$ is the *Geometric realization* functor. Now we can repeat the same procedure with Path 1-Groupoid functor $\pi'_{\leq 1}: Man \rightarrow Groupoids$.

**My questions are the following:**

Is $|N \circ \pi_{\leq 1}(X)|= |N \circ \pi'_{\leq 1}(X)|$? (where "$=$" is in an appropriate sense)

Is there a way to present a Path $\infty$-groupoid of a smooth space such that it is different from Smooth Fundamental $\infty$-groupoid of the space? (So that it matches our intuition for $n=1$ case)

(By "$n$" I mean "Groupoids in the level 1").