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I need a - I suppose - classical result, but whose reference is difficult to find:

Given $M^2$ and $N^1$ two smooth manifolds, every continuous embedding $h \colon N \hookrightarrow M$ is homotopic to a smooth embedding, that is arbitrary closed to $h$. If the manifolds have boundary, we suppose that $h^{-1}(\partial M) = \partial N$.

I think that is a classical result, dating back to the last century. I had found a reference, that I've lost. Does somebody have a nice reference? (Obviously supposing that I've written a true theorem).

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  • $\begingroup$ I think it is in Hirsch, Differential Topology. $\endgroup$ Commented Jun 12 at 12:17
  • $\begingroup$ I think the theorems in Hirsch are all quite general, and do not handle cases in specific small dimensions. I assume OP intends by the superscripts that $\dim M = 2$ and $\dim N = 1$, as in general this is quite false. $\endgroup$ Commented Jun 13 at 0:36
  • $\begingroup$ In the absence of a reference, would a proof be OK? $\endgroup$ Commented Jun 15 at 0:43
  • $\begingroup$ Yea, it would be fine. (As tou said, I'm supposing $\dim M = 2$ and $\dim N = 1$ $\endgroup$ Commented Jun 19 at 16:11

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I don't know a reference but here is a short proof, provided a reference to some other facts. Please let me know if I should explain some details more carefully or work harder to get a reference; I am sure a concise reference for Fact 1 exists.

Your result follows from the combination of the two facts stated below, and the fact that isotopies preserve the property of being an embedded submanifold.

Fact 1. There exists a smooth surface $M'$, a smooth submanifold $N' \subset M'$, and a homeomorphism $h: M' \to M$ for which $h(N') = N$.

Fact 2. Any homeomorphism between smooth surfaces may be isotopied to a diffeomorphism by an arbitrarily small isotopy.

Proof of Fact 1. I will assume $N'$ is two-sided, as the argument is a little bit more awkard if it isn't. Cut open $M$ along $N$ to obtain a compact surface $M_0$, possibly with corners, so that pasting together part of the boundary of $M_0$ results in $M$. Because every surface has a smooth structure, so too does $M_0$. By an isotopy of the gluing map, one may assume that the pasting operation on $\partial M_0$ is smooth (if $N'$ has one-sided components, you need to know that every homeomorphism $\iota: S^1 \to S^1$ with $\iota^2 = 1$ is conjugate to a smooth one). Performing this gluing gives you $M'$, with $N'$ the image of the boundary curves.

Proof of Fact 2. Hatcher gives a nice exposition of this fact here (https://arxiv.org/abs/1312.3518). The fact that the isotopy can be made small is mentioned on page 7.

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  • $\begingroup$ Thanks, perfect! $\endgroup$ Commented Aug 21 at 15:10

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