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Let $M,N$ be smooth manifolds. It seems to be well known that if the sets $C^0(M,N)$ and $C^\infty(M,N)$ are equipped with the appropriate topologies (I suppose the weak/strong Whitney topology), then the inclusion $$C^\infty(M,N) \hookrightarrow C^0(M,N)$$ is a weak homotopy equivalence, see e.g. here. However, I don't know of any reference where this is proved explicitly.

Can anyone provide a reference that the above inclusion is a weak homotopy equivalence?

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2 Answers 2

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At most this requires the relative smoothing theorem: if $f: X \to N$ is $C^0$, and smooth on some closed subset $M$, then it is homotopic to a smooth map, with $f|_M$ fixed by this homotopy. Apply this to $X = S^n \times M$ to see that the map is a surjection on homotopy groups. (That is, we're smoothing a "sphere's worth" of continuous maps.) Now apply a version of this theorem with boundary on $S^n \times M \times I$ to see that it's an injection on homotopy groups. All of these theorems are proved in Hirsch, differential topology.

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  • $\begingroup$ As this answer contains the key idea to prove the weak homotopy equivalence for $C^\infty(M,N)$ and $C(M,N)$ equipped with the weak (sometimes called compact-open) topology I have decided to accept the answer. There are however some remarks that should be made: 1) The answer relies on the fact that for compact Hausdorff spaces $K$ (such as $S^n$ and $S^n \times I$) a continuous map $K \to C^0(M,N)$ is exactly the same as a continuous map $K \times M \to N$. 2) The inclusion $C^\infty(M,N) \hookrightarrow C(M,N)$ is no longer a weak equivalence when the spaces are equipped with the strong $\endgroup$
    – Adrian Clough
    May 27, 2016 at 21:03
  • $\begingroup$ (sometimes called Whitney) topology, when $M$ is not compact. This follows from the fact that in this topology a map $I \to C^0(M,N)$ corresponds to a homotopy which is constant on all but a compact subsets of $M$. See the answer to this question. 3) Up to weak homotopy equivalence these two topologies seem to be the the only ones considered. See section 5.1 of these lecture notes. $\endgroup$
    – Adrian Clough
    May 27, 2016 at 21:10
  • $\begingroup$ @AdrianClough Yes, I'm sorry this wasn't in the answer. I had only intended to consider the case when $M$ is compact, when these topologies agree. Thanks for pointing out the subtleties when $M$ is not compact. $\endgroup$
    – user98602
    May 27, 2016 at 21:16
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You can find the detailed proof in my joint paper with Olexandra Khokhliyuk for the case of weak Whitney topology:

[1] O. Khokhliuk, S. Maksymenko, Smooth approximations and their applications to homotopy types. Proceedings of the International Geometry Center, vol. 13, no. 2 (2020) 68-108 http://doi.org/10.15673/tmgc.v13i2.1781 or https://arxiv.org/pdf/2008.11991.pdf

I needed this kind of results two years ago and searched for the proof but failed to find it. Finally we decided to write the detailed proof. The paper also contains variants for manifolds with corners, inclusions of open sets (as mentioned by Ryan Budney here), and relative variant for maps with fixed restriction to some subset. It also contains references to partial results and some posts in the internet including the present post by Adrian Clough.

When we put the preliminary version of the paper to arXiv, Helge Glockner informed us that such a statement follows from his (currently yet unpublished) paper

[2] Helge Glockner. Homotopy groups of ascending unions of infinite-dimensional manifolds (2008) https://arxiv.org/pdf/0812.4713.pdf

Then we also included (in Remark 7.5 of [1]) explanations of how to get the mentioned result from Helge Glockner's paper [2].

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