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I am not an expert in Differential Topology, so let me apologize if this question admits a straightforward answer. I checked some standard references, but I could not find one.

Let $M$ be a smooth $n$-manifold with boundary $N:= \partial M$, and assume that there exists a continuous retraction of $M$ onto $N$, namely, a continuous map $r \colon M \to N$ that is the identity on $N$.

Question. Does there exist a smooth retraction $s \colon M \to N$? And what about a smooth retraction homotopic to $r$?

I am aware that, by Whitney Approximation Theorem, $r \colon M \to N$ is homotopic to a smooth map, but I do not see how to conclude from this that it is homotopic to a smooth retraction.

Motivation. This question arose when I was trying to mimic the standard proof of Brouwer Fixed Point Theorem (for continuous maps $f \colon \mathbb{D}^n \to \mathbb{D}^n$) by using de Rham cohomology, instead of singular homology. In the homology setting, the starting point is the non-existence of a continuous retraction $r \colon \mathbb{D}^n \to S^{n-1}$, which is obtained by contradiction looking at the functorial group homomorphisms induced by $r$ and by the inclusion $i \colon S^{n-1} \to \mathbb{D}^n$. But de Rham cohomology is only functorial with respect to smooth maps...

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    $\begingroup$ There is a relative version of the approximation result you mentioned: any continuous map $f\colon M\rightarrow M'$ that is smooth on some closed subset $A\subset M$ is homotopic to a smooth map via a homotopy that is constant on $A$. $\endgroup$ Nov 4, 2022 at 7:42
  • $\begingroup$ by "that is constant on $A$" did you mean "that is the identity on $A$?" Or "that coincides with $f$ on $A$?" $\endgroup$ Nov 4, 2022 at 7:56
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    $\begingroup$ The latter. This should be in Hirsch’ for example $\endgroup$
    – Thomas Rot
    Nov 4, 2022 at 8:02
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    $\begingroup$ @FrancescoPolizzi the proof of Brower's theorem by approximation is a slightly more complicated than that, see pg.14-15 of Milnor. Topology from a differentiable viewpoint. University Press of Virginia. 1965. $\endgroup$ Nov 5, 2022 at 0:58
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    $\begingroup$ The argument in Milnor’s book is what I had in mind. $\endgroup$ Nov 5, 2022 at 1:19

1 Answer 1

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Using a collar of the boundary we resort to the case when $r\in C^0(M, \partial M)$ is given by the projection $\partial M\times I \to \partial M$ over the collar .

Since $r$ is smooth on an open neighbourhood of $\partial M$, for any $\epsilon>0$ we can find a smooth $g\in C^\infty(M,\partial M)$ such that $g=r$ in a neighbourhood of $\partial M$ and $|g-r|<\epsilon$.

For a reference see Prop. 3.11 of Milnor's lectures on Differential Topology 1958 (notes taken by James Munkres).

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