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Let $X$ and $Y$ be topological spaces. A homotopy from a continuous function $f: X \to Y$ to a continuous function $g: X \to Y$ is a continuous function $H : X \times [0, 1] \to Y$ such that $H(x, 0) = f(x)$ and $H(x, 1) = g(x)$ for all $x \in X$. An isotopy is a homotopy such that $H(\cdot, t)$ is an embedding for every $t \in [0, 1]$; isotopy requires $f$ and $g$ to be embeddings.

Suppose $X = \{x \in \mathbb{R}^n : \sum_{i = 1}^n x_i^2 < 1\}$ is the open $n$-ball, and let $f, g : X \to Y$ be homeomorphisms. Does there always exist an isotopy from $f$ to $g$?

Alexander's trick solves a similar problem, but on closed $n$-balls.

kaba
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Yes, but it is not as elementary as the Alexander trick. Your group is the same as the space of homeomorphisms of $\Bbb R^n$, and you want to know that the subspace of orientation-preserving homeomorphisms is connected. The case $n=1$ is easy to prove by hand, so we will suppose $n>1$.

In fact, $\text{Homeo}(\Bbb R^n) \cong \text{Homeo}(S^n, \text{pt})$, the space of homeomorphisms of the sphere that fix a point. This sits into a fibration $$\text{Homeo}(S^n, \text{pt}) \to \text{Homeo}(S^n) \to S^n,$$ and similarly for the orientation-preserving versions (which I denote $\text{Homeo}^+$).

So if $\text{Homeo}^+(S^n)$ is connected, we see from the long exact sequence $$\cdots \to \pi_1 S^n \to \pi_0 \text{Homeo}^+(S^n, \text{pt}) \to \pi_0 \text{Homeo}^+(S^n) \to \pi_0 S^n$$ that $\pi_0 \text{Homeo}^+(\Bbb R^n)$ is connected.

What remains is an inductive argument using the annulus theorem to show that this group is connected; I have written a proof out in the second half of this answer. The idea is to first demand your homeomorphism sends the equator to itself, inductively isotope that to the identity, and then use the Alexander trick on the two hemispheres.

  • How do you define an orientation-preserving homeomorphism? – kaba Feb 02 '19 at 19:56
  • You define orientations homologically. –  Feb 02 '19 at 20:03
  • The link uses the notation $H_n(M, M \setminus {p}; \mathbf{Z})$. It looks like a relative singular homology group, but what is the meaning of $\mathbf{Z}$? – kaba Feb 02 '19 at 22:01
  • @kaba the integers. –  Feb 02 '19 at 22:26
  • What I mean is that I have not seen a three-argument form of a homology group before. What is the role of the third argument? – kaba Feb 03 '19 at 08:56
  • Perhaps the third argument is a ring (?) for the coefficients of the singular simplices, which usually is the integers, but can be generalized? – kaba Feb 03 '19 at 09:14
  • Yes, this is correct (but you only need an abelian group). I don't really know sources that don't simultaneously introduce homology with arbitrary coefficients. Certainly this is what Hatcher does. –  Feb 03 '19 at 13:38
  • If this is confusing you, there is almost no harm in dropping the + and modifying the inductive argument to see that $\pi_0 \text{Homeo}(\Bbb R^n) = \Bbb Z/2$ more directly. The only difficulty is showing that reflection on a hyperplane is not isotopic to the identity. This requires the notion of orientation. –  Feb 03 '19 at 16:58
  • What I think I understood up to this point: 1) the homeomorphism $\text{Homeo}(\mathbb{R}^n) \approx \text{Homeo}(S^n, pt)$ follows from one-point compactification. 2) The fibration is a Serre fibration. 3) The long exact sequence holds for Serre fibrations in general. 4) There is a typo in "$\pi_0 \text{Homeo}^+(\mathbb{R}^n)$ is connected."; $\pi_0$ should be removed. 5) "show that this group is connected" refers to $\text{Homeo}^+(S^n)$. Therefore, what is left is to understand the annulus theorem. – kaba Feb 03 '19 at 20:03
  • Thanks for catching the typo. (I won't edit to avoid bumping). I do not suggest trying to understand the annulus theorem. It is exceedingly difficult topology in dimensions greater than two, and in dimension 2 is similar to the Schoenflies theorem. This is one of those things that is better left as a black box unless you are trying to extend it. –  Feb 03 '19 at 20:06
  • Sorry, I meant what is left is to understand your proof which uses the annulus theorem. I'll dive into that next. – kaba Feb 03 '19 at 20:10
  • Everything seems to check out! The final thing that remains unclear to me is orientation. What I think has been shown is that if I have two orientation-preserving homeomorphisms of $X$, then they are in the same connected component. However, if one of the homeomorphisms is not orientation-preserving, then they are in different components, and there is no isotopy, right? – kaba Feb 03 '19 at 21:02
  • @kaba Correct. I do not see an elementary (non-homological) argument for this, but it's frankly surprising we got this far at all while never mentioning homology. :) –  Feb 03 '19 at 23:07