Let $M$ be a manifold of dimension $d$ and let $\mathsf{Disk}_{/M}$ be the category of open subsets of $M$ that are diffeomorphic to $\mathbb{R}^d$ with morphisms given by inclusions. Let $\mathrm{B} \mathsf{Disk}_{/M}$ be the classifying space of this category. How do I prove that $\mathrm{B} \mathsf{Disk}_{/M}$ is homotopy equivalent to $M$? Intuitively, $\mathrm{B} \mathsf{Disk}_{/M}$ should be some thickening of $M$.
I think this might follow from Quillen's theorem A. In order to apply it, I would first need to exhibit $M$ as the classifying space of some more tractable category. I thought about $M \simeq \left\lvert \operatorname{Sing} M \right\rvert$, but $\operatorname{Sing} M$ is not the nerve of a category. The second barycentric subdivision of $\operatorname{Sing} M$ is a category, but it is pretty complicated.
Alternatively, we might try a direct argument. If I'm not mistaken, we have $$\mathrm{B} \mathsf{Disk}_{/M} \simeq \operatorname*{colim}_{[n] \to \mathsf{Disk}_{/M}} \Delta^n.$$ On the other hand, $$M \cong \operatorname*{colim}_{\mathsf{Disk}_{/M}} \mathbb{R}^d \simeq \operatorname*{hocolim}_{[0] \to \mathsf{Disk}_{/M}} \Delta^0.$$ So one could try to prove that the index categories $\mathsf{Disk}_{/M}^{\mathbf{\Delta}}$ and $\mathsf{Disk}_{/M} \cong \mathsf{Disk}_{/M}^{[0]}$ are sufficiently similar. I don't quite know what this means precisely, though.
I'll also be happy if someone could provide a reference for the proof - it certainly ought to be well-known.
Edit: Writing $M \simeq \operatorname*{hocolim}_{[0] \to \mathsf{Disk}_{/M}} \Delta^0$ is a little duplicitous. In the first colimit for $\mathrm{B}\mathsf{Disk}_{/M}$, the gluing data is specified by the simplicial identities in $\mathbf{\Delta}$, but in the second homotopy colimit, the gluing data is not visible at all in the expression - it depends on how all the various $\mathbb{R}^d$'s intersect - and is obscured by writing $\mathbb{R}^d \simeq \Delta^0$, which superficially cannot intersect meaningfully. In other words, this makes the (possible) comparison less obvious.
I think I have at least a sketch of an argument. Recall, every manifold admits a locally finite good open cover $\mathcal{U}$. Let us write $\mathcal{U}$ also for the poset category of nonempty finite intersections generated by elements of $\mathcal{U}$ ordered by inclusion. The nerve theorem in this case should apply and we deduce that $\mathrm{B}\mathcal{U} \simeq M$.
There is an inclusion of categories $\mathcal{U} \hookrightarrow \mathsf{Disk}_{/M}$. We want to apply Quillen's theorem A to this to show $\mathrm{B}\mathcal{U} \to \mathrm{B}\mathsf{Disk}_{/M}$ is a homotopy equivalence, and then we would be done. To apply the theorem, we need to check that for each $\mathbb{R}^d$ in $\mathsf{Disk}_{/M}$, the comma category $\mathcal{U}_{\mathbb{R}^d/} := \mathcal{U} \times_{\mathsf{Disk}_{/M}} \mathsf{Disk}_{\mathbb{R}^d/ /M}$ is empty or contractible. But $\mathcal{U}_{\mathbb{R}^d/}$ has an initial object, namely the intersection of all open sets in the cover containing $\mathbb{R}^d$, hence it is contractible if not empty. This completes the argument.