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The definition of regular homotopy from Wikipedia says that two immersion $f,g:M\to N$ are regularly homotopic if they represent points in the same path-component of $\text{Imm}(M,N)$.

What does "represent points in the same path-component" mean? Is there an accurate Math explanation?

It also says that regular homotopy is similar to isotopy of embedding, but they are not the same. From Smale's Paradox, we know there is a regular homtopy between $f,-f:\mathbb S^2\to\mathbb R^3$, but there is no isotopy between $f$ and $-f$ because ambient isotopy keeps orientation.

What is the geometric difference between regular homotopy and isotopy of embedding by taking $f,-f:\mathbb S^2\to\mathbb R^3$ as an example? Here $f$ is the standard embedding of $\mathbb S^2$.

Any advice is helpful. Thank you.

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1 Answer 1

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If $f$ and $g$ are in the same path component of $\operatorname{Imm}(M, N)$, then there is a path $p : [0, 1] \to \operatorname{Imm}(M, N)$ such that $p(0) = f$ and $p(1) = g$; i.e. $p$ is a path connecting $f$ and $g$. Now consider $H : M \times [0, 1] \to N$ given by $H(x, t) = p(t)(x)$, then $H(x, 0) = p(0)(x) = f(x)$ and $H(x, 1) = p(1)(x) = g(x)$ and for any $t \in [0, 1]$, $H(x, t) = p(t)(x)$ with $p(t) \in \operatorname{Imm}(M, N)$; also note that such a homotopy defines a path in $\operatorname{Imm}(M, N)$ between $f$ and $g$. So $f, g \in \operatorname{Imm}(M, N)$ are regularly homotopic if and only if they are homotopic via immersions (i.e. for every $t \in [0, 1]$, $H(\cdot, t)$ is an immersion) - this is usually taken as the definition of regularly homotopic. Note, this is a stronger condition than saying $f$ and $g$ are homotopic as the latter has no restrictions on the intermediate maps.

More generally, if $C(M, N)$ denotes the space of continuous maps from $M$ to $N$, and $\operatorname{Emb}(M, N)$ the space of embeddings of $M$ into $N$, then

  • $f, g \in C(M, N)$ are homotopic if and only if they are in the same path-connected component of $C(M, N)$,
  • $f, g \in \operatorname{Imm}(M, N)$ are regularly homotopic if and only if they are in the same path-connected component of $\operatorname{Imm}(M, N)$,
  • $f, g \in \operatorname{Emb}(M, N)$ are isotopic if and only if they are in the same path-connected component of $\operatorname{Emb}(M, N)$.

As $\operatorname{Emb}(M, N) \subseteq \operatorname{Imm}(M, N) \subseteq C(M, N)$, one sees immediately that if $f, g \in \operatorname{Emb}(M, N)$ are isotopic, then they are regularly homotopic and homotopic; these implications can also be observed directly from the definitions.

The geometric difference between a regular homotopy and an isotopy is the same as the geometric difference between an immersion and an embedding. Every embedding is an immersion but not vice verca (i.e. $\operatorname{Emb}(M, N)$ is usually a proper subset of $\operatorname{Imm}(M, N)$). At some point in the regular homotopy, one of the intermediate maps will be an immersion but not an embedding; for example, whenever the image of the sphere has self-intersection. One of the explicit regular homotopies between $f$ and $-f$ was given by Morin; halfway through the regular homotopy, the image of the sphere is Morin's surface (image below from Wikipedia) which has self-intersection, so it cannot be the image of an embedding.

$\hspace{26mm}$enter image description here

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  • $\begingroup$ Clear and detailed. Thank you. $\endgroup$
    – gaoxinge
    Jan 8, 2015 at 0:56
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    $\begingroup$ I apologise for excavating this question from years ago, but your explanation is outstanding. May I ask what topology you're giving to the space of immersions $\mathrm{Imm}(M, N)$? Compact-open? $\endgroup$
    – Acton
    Aug 19, 2018 at 3:39
  • $\begingroup$ Yes, I think compact-open is the relevant topology. $\endgroup$ Aug 19, 2018 at 4:31

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