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I have some problems with homotopies.

The situation is this:

Let $X$ be a surface, which is homeomorphic to a 2-Sphere with a finite number (at least 3) of points removed (equivalently, an open Annulus with a finitely many punctures).

$f: X \rightarrow X$ is a homeomorphism, which is isotopic to the identity.

Now,

let's say we have two isotopies $\tilde{f},\tilde{g}: [0,1]\times X \rightarrow X$ of $f$ to the identity, with $$ \tilde{f}(0,x) = \tilde{g}(0,x) = x \quad \forall x \in X,\\ \tilde{f}(1,x) = \tilde{g}(1,x) = f(x) \quad \forall x \in X. $$

Let $x$ be a point in $X$. We can now look at two paths $\alpha, \beta : [0,1] \rightarrow X$, definded by $$ \alpha(t) = \tilde{f}(t,x),\\ \beta(t) = \tilde{g}(t,x). $$

The two paths have the same start- and endpoints, namely $x$ and $f(x)$. It seems to be a well-known fact (I read about it in "Farb, Margalit - A primer on mapping class groups (p.43)"), that on this surface, the maps $\tilde{f}$ and $\tilde{g}$ are homotopic. Thus, there exists a continuous map $h:[0,1]\times [0,1] \times X \rightarrow X$, with $$ h(0,t,x) = \tilde{f}(t,x),\\ h(1,t,x) = \tilde{g}(t,x), \quad \forall t \in [0,1], \forall x \in X. $$

It is clear now, that $h$ gives us a free homotopy of the two paths $\alpha$ and $\beta$, by sending $(s,t) \mapsto h(s,t,x)$ but I don't see a reason, why $\alpha$ and $\beta$ should be homotopic in the usual sense of a homotopy between paths (i.e. fixing the endpoints). My problem is, that this is exactly what is claimed to be true in a paper by John Franks (page 4).

Also, on the open annulus $A = (0,1) \times \mathbb{R}/ \mathbb{Z}$, this is clearly not true: Let $f$ be the identity on $A$. Then, there are two isotopies $\tilde{f}$ and $\tilde{g}$ of $f$ to itself: $$ \tilde{f}(t,[x]) = [x + t],\\ \tilde{g}(t,[x]) = [x], $$ where $[x]$ is an equivalence class in $\mathbb{R}/ \mathbb{Z}$. Now, $\alpha$ is a loop that winds around the hole in the center and $\beta$ is the constant loop. These two are clearly not homotopic in the sense of a path-homotopy, so the argument has to stem somehow from the number of punctures... I'm confused. Any help is much appreciated!

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The special feature of $X$, a sphere with three or more punctures, that is being used here is that the space $E(X)$ of all homotopy equivalences $X\to X$ has $\pi_1 E(X)=0$. (Here we take the identity map of $X$ as the basepoint of $E(X)$ for computing $\pi_1 E(X)$.) The corresponding statement when $X$ is an annulus is not true, since $\pi_1 E(X)={\mathbb Z}$ in this case. Granted that $\pi_1E(X)=0$, this implies that your two paths $\tilde f$ and $\tilde g$ in $E(X)$ are homotopic in the usual sense of homotopy of paths, i.e., as paths in $E(X)$ joining the identity map to $f$. This homotopy from $\tilde f$ to $\tilde g$ then gives a homotopy from $\alpha$ to $\beta$ through paths from $x$ to $f(x)$.

One can prove that $E(X)$ is simply-connected as follows. By evaluating each homotopy equivalence $X\to X$ at a basepoint $x_0\in X$ we obtain a map $p:E(X)\to X$. This is a fibration and its fiber over $x_0$ is the space $E(X,x_0)$ of basepoint-preserving homotopy equivalences $X\to X$. Since $X$ is a $K(\pi,1)$, the components of $E(X,x_0)$ have trivial homotopy groups, and in particular are simply-connected. Part of the long exact sequence of homotopy groups for the fibration is $$\pi_1 E(X,x_0)\to \pi_1 E(X) \to \pi_1 X \to \pi_0 E(X,x_0)$$ The last group here is isomorphic to the automorphism group of $\pi_1X$ by sending a basepoint-preserving homotopy equivalence to the induced automorphism of $\pi_1X$. The third map in the sequence sends an element of $\pi_1 X$ to the inner automorphism given by conjugation by that element. Thus the kernel of the third map is the center of $\pi_1 X$. When $X$ is a sphere with 3 or more punctures, $\pi_1 X$ is free on 2 or more generators so its center is trivial. Exactness then forces $\pi_1 E(X)$ to be 0 since the first term in the sequence is 0.

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  • $\begingroup$ Thanks alot! "The third map in the sequence sends an element of $\pi_1 X$ to the inner automorphism given by conjugation by that element." Is there a quick way to see why this is true? $\endgroup$ – herbert west Jan 7 '14 at 9:34

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