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As it is well-known (and for example this question shows) each continuous map between smooth manifolds is homotopically equivalent to a smooth map that can be constructed using the Whitney embedding theorem.

My question is: does an intrinsic proof of this result exist, i.e. one that avoids using Whitney or other embedding theorems?

My vague idea is the following: without loss of generality, we can assume that the smooth manifolds are Riemannian manifolds. Now I would like to say something along the lines of "since we can smooth our continuous function locally, we can try to exploit the metric to define a notion of convex combination of points on the codomain manifold so that we can attach the local pieces together using a partition of unity in the same way as we do in the Euclidean case". The idea seems nice, but I suspect that it might not work because convex combination of points might be hard to define in a smooth way, but I don't know.

A similar and very related question is how to show that the collection of smooth maps between two Riemannian manifolds is dense in the set of continuous maps w.r.t. the sup norm without resorting to the Nash embedding theorem.

Every comment is very much appreciated.

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  • $\begingroup$ Aren't smooth functions and continuous functions both homotopy sheaves? If so, since the inclusion $C^\infty (U,N) \rightarrow C^0 (U,N)$ is a weak equivalence when $U \cong \mathbb{R}^n$, it should follow that $C^\infty (M,N) \rightarrow C^0 (M,N)$ is a weak equivalence. $\endgroup$ Commented Oct 24, 2022 at 16:24
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    $\begingroup$ @ConnorMalin I am no expert of sheaves but I suspect that to prove that the inclusion is a weak equivalence one needs to already know that continuous maps can be smoothed. $\endgroup$
    – No-one
    Commented Oct 25, 2022 at 8:55
  • $\begingroup$ See Hirsch's Differential Topology: people.dm.unipi.it/benedett/HIRSCH.pdf $\endgroup$ Commented Oct 25, 2022 at 9:19
  • $\begingroup$ Or Kosinski's Differential Manifolds: maths.ed.ac.uk/~v1ranick/papers/kosinski.pdf $\endgroup$ Commented Oct 25, 2022 at 9:25
  • $\begingroup$ @Titti Both of these spaces are contractible since $\mathbb{R}^n$ is smoothly contractible. $\endgroup$ Commented Oct 25, 2022 at 15:27

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