5
$\begingroup$

Does the space of homotopy equivalences of $S^1$ deformation retract onto the space of homeomorphisms of $S^1$? If so, does anyone have a reference?

I found that Kneser proved that $Homeo(S^1)$ deformation retracts onto $O(2)$ and $Homeo^+(S^1)$ deformation retracts onto $SO(2)$ (orientation preserving homeos deformation retracts onto rotations). I'd like the space $HE^+(S^1)$ of degree 1 homotopy equivlances of $S^1$ to deformation retract onto these. The space I'm calling $HE^+(S^1)$ may go by $HomEq(S^1)$ or $SG_n$ and seems to be of interest to homotopy theorists for higher $n$.

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ Yes, it does. The proof doesn't really require a reference as it's a straight-line homotopy argument, using only the $\mathbb R \to S^1$ model for the covering space. I'm pretty sure this is discussed in many textbooks. Thurston's Geometry and Topology textbook discusses this immediately before Smale's theorem about $Diff(S^2)$. $\endgroup$
    – Ryan Budney
    Commented May 12, 2011 at 18:15
  • $\begingroup$ Thank you for the prompt responses. I think Hatcher's reply to "Homotopy type of set of self homotopy equivalences of a surface" might be more helpful. Both of the replies here seem to only address connectivity of components of HE, i.e. that any homotopy equivalence is homotopic to a homeomorphism. Of course this is the first step, but one needs to address the contractibility of the fibers HE to S^1 which apparently comes from the fact that S^1 is a K(\mathbb{Z}, 1) $\endgroup$
    – Aaron Magid
    Commented May 12, 2011 at 19:45

2 Answers 2

Reset to default
3
$\begingroup$

Put $$HE^+_1(S^1)=\{f\in HE^+(S^1):f(1)=1\}. $$
There is an evident homeomorphism $m:S^1\times HE_1^+(S^1)\to HE^+(S^1)$ given by $m(z,f)(x)=z f(x)$, and this restricts to give a homeomorphism $S^1\times Homeo_1^+(S^1)\to Homeo^+(S^1)$.

It will thus suffice to discuss $HE_1^+(S^1)$ and $Homeo_1^+(S^1)$.

Next, define $e:\mathbb{R}\to S^1$ by $e(t)=\exp(2\pi i t)$. Let $X$ denote the space of maps $$ u:[0,1]\to\mathbb{R} $$ with $u(0)=0$ and $u(1)=1$, and let $Y$ be the subspace of strictly increasing maps. For any $u\in X$ there is a unique map $p(u):S^1\to S^1$ with $p(u)(e(t))=e(u(t))$ for all $t\in [0,1]$. This construction gives a homeomorphism $p:X\to HE^+_1(S^1)$, and restricts to a homeomorphism $p:Y\to Homeo_1^+(S^1)$. This is a fairly straightforward exercise with covering space theory and topologies on mapping spaces. Now $X$ and $Y$ are both convex, so they have obvious contractions to the identity map given by $$ h(t,u)(x) = tx+(1-t)u(x). $$

It is not hard to see that $Y$ is not closed in $X$, so it cannot be a retract, so $Homeo_1^+(S^1)$ is not a retract of $HE_1^+(S^1)$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ So to clarify, the answer to my original question is negative. Homeo is not a deformation retract of HE because it isn't closed, but both deformation retract onto O(2) and both your answer and Agol above give fairly explicity ways of seeing this. Thanks to all for the help. $\endgroup$
    – Aaron Magid
    Commented May 12, 2011 at 21:33
3
$\begingroup$

I think Ryan Budney's comment can be made to work. In order to take a straight line homotopy, you need to figure out which rotation you are going to homotope to, and make this choice in a continuous fashion. Here's one possible choice. For a homotopy equivalence $f:S^1\to S^1$, take a lift $F:\mathbb{R}\to\mathbb{R}$. Then $F$ has the property that $F(x+n)=F(x)+n, n\in \mathbb{Z}$ ($\mathbb{Z}$-equivariant). The function $F(x)-x$ is therefore periodic, and attains a minimal value $m(f)$. Take the straightline homotopy from $F(x)$ to $x+m(f)$. This homotopes the continuous $\mathbb{Z}$-equivariant functions to the functions of the form $x+r, r\in\mathbb{R}$, and therefore descends to a homotopy of $HE^+(S^1)$ to $SO(2)$. I think it's clear that this homotopy is continuous with respect to the topology on $HE^+(S^1)$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thanks Ian. I believe this is continuous with respect to the topology on HE. You're right that this shows HE retracts to SO(2), but I think Neil is correct to point out I was wrong in my original question since Homeo isn't closed in HE. $\endgroup$
    – Aaron Magid
    Commented May 12, 2011 at 21:37
  • $\begingroup$ Woops, I should have read the question more carefully! This gives a homotopy to SO(2), but obvious not a deformation retract to $Homeo^+(S^1)$. $\endgroup$
    – Ian Agol
    Commented May 12, 2011 at 22:12
  • $\begingroup$ There may be a way to homotope $HE^+(S^1)$ to "monotonic" maps. For a $\mathbb{Z}$-equivariant map $F:\mathbb{R}\to \mathbb{R}$, define $F'(x)= \max\\{ F(y) | y\leq x\\}$. Then you could straightline homotopy from $F$ to $F'$, maintaining that the map is $\mathbb{Z}$-equivariant. Intuitively, this takes the graph of $F$ and "fills in" the "valleys" to get a monotonic map. This should give a deformation retract to monotonic maps. $\endgroup$
    – Ian Agol
    Commented May 12, 2011 at 22:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .