Suppose that $M$ is a differentiable space in the sense of Chen (cf. https://ncatlab.org/nlab/show/Chen+space ).

Assume that $M$ also has the structure of a topological space and that the two structures are compatible in the sense that every plot is a continuous map.

To distinguish the structures, let ${}_SM$ denote the differentiable structure and let $M_T$ denote the topological one. There is an adjoint pair $$ L: \text{Differentiable Spaces} \leftrightarrows \text{Topological Spaces}: R $$ in which $L(X)$ is $X$ as a set with the topology which defines a subset $O\subset X$ to be open iff its preimage with respect to every plot is open. The functor $R$ gives a space $Y$ a smooth structure in which every continuous map $U \to Y$ from a convex set in euclidean space is declared to be a plot.

With $M$ as above, we have a continuous map $L ({}_{S}M)\to M_T$.

Hypothesis: The map $L ({}_{S}M)\to M_T$ is a weak homotopy equivalence.

(I.e.: every continuous map from a sphere to $M_T$ can be deformed to a smooth map and every continuous map from a disk to $M_T$ which is smooth on its boundary can be deformed rel boundary to a smooth map.)

Let $$ \text{plot}_M $$ be the category of plots of $M$. An object of this category is a plot $\phi: U \to M$. A morphism $(U,\phi) \to (V,\psi)$ is a $C^\infty$ map $f: U\to V$ such that $\psi \circ f = \phi$. Then there is an evident map $$ \underset{(U,\phi)}{\text{colim } } U \to M_T . $$

Question: Under what reasonable conditions can we conclude that this map is a weak equivalence?

Note: I really do want the colimit here (not the homotopy colimit).

Here is a some evidence: Let $g: S^k \to M_T$ be continuous. By my hypothesis, we can assume that $g$ is smooth, i.e., we can assume $g: S^k \to {}_SM$ is a smooth map (i.e., it preserves plots where $S^k$ is given its usual differentiable structure). Then we have a commutative diagram $\require{AMScd}$

\begin{CD} \underset{\phi: U \to M}{\text{colim } } U @>>> M \\ @AAA @AAA \\ \underset{\phi: U \to S^k}{\text{colim } } U @>>> S^k \end{CD}

But the bottom map is a retraction up to homotopy: use a smooth triangulation of $S^k$ to construct a section up to homotopy. This shows that top horizontal map is a surjection on homotopy.


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