Recently I read on the nlab, that the category $\mathsf{SmoothMf}$ of smooth manifolds can be realized as the Karoubi-envelope / Cauchy-completion of the category $\mathsf{SmoothOpen}$ of open subsets of euclidean spaces $\mathbb R^n$ and smooth maps between them. The proof invokes the tubular neighborhood theorem, which (according to a rather coarse search of the internet) seems to be a special property of $\mathsf{SmoothMf}$.
Regardless I am wondering, whether we can identify the category of topological manifolds $\mathsf{TopMf}$ with the Cauchy-completion of $\mathsf{TopOpen}$, using continuous maps instead of smooth ones.
Similarly, is the (or some nice subcategory, say smooth ones) category of schemes $\mathsf{Sch}$ realizable as the Cauchy-completion of the category $\mathsf{Aff} = \mathsf{CRing}^{op}$ of affine schemes?
I feel like this question is natural to ask, so it should be answered already, but didn't find any positive nor negative results. Unfortunately I don't really know enough about the differences between $\mathsf{SmoothMf}$ and $\mathsf{TopMf}$ nor do I know much about Cauchy-completions. So I apologize, if there are more than obvious obstructions I just didn't notice yet. Anyway, thank you for your time.
No. The Cauchy completion of open subsets of $\def\R{{\bf R}} \R^n$ with continuous maps (instead of smooth) is much bigger than the category of topological manifolds.
For example, it contains the gluing of three copies of $[0,∞)$ along $\{0\}$. It is easy to construct it as a retract of $\R^2$, using a continuous map that (say) projects in the direction of a fixed ray in each of the three regions of $\R^2$. This map is not smooth.
The reason this does not work in the smooth category is that the retraction must be differentiable at the point $0$, so the derivative must have rank at least 2, which means the retract cannot be 1-dimensional.