space of homotopy equivalences of $S^1$ 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 $n$.
 A: 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 $HE^+(S^1)$. 
A: 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 $HE_1^+(S^1)$.  
