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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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  • $\begingroup$ first we are fixing a point $x\in X$ and then defining $\alpha$ and $\beta,$ which gives us that both $\alpha$ and $\beta$ start at $x$ and end at $f(x).$ So, is the question like, for the sphere minus a finite set of points case, how are these two paths homotopic (fixing end-points), if so then it should follow from the way we define isotopy of homeomorphisms. What I mean to say is since $\tilde{f}$ and $\tilde{g}$ are isotopic then it will follow right? $\endgroup$ – Vidit D Dec 29 '18 at 17:43
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I can at least answer why it fails for the annulus.

Your example of the annulus is homotopically equivalent to $S^1$ which has an Euler characteristic of $0$. Your source (John Franks, Definition 2.1, page 4) stipulates that the space have negative Euler characteristic. This means that it's the sphere with at least 3 holes punched into it, as each hole decreases the Euler characteristic by $1$.

I believe the problem arises for the annulus because the space of homeomorphisms is not simply connected. Imagine the homeomorphisms of the annulus: apart from weird internal warpings and whatnot, this space definitely has a loop in it somewhere because we can perform rotations of the annulus about the origin by any angle we like. A $2 \pi$ rotation is the same as the identity, but there are definitely homotopically distinct isotopies.

Let me give an example. Define $R(\theta)$ to be a rotation of the annulus by $\theta$ (a homeomorphism). Define the set of isotopies parametrised by integer $n$ as $$\tilde f_n : [0,1] \rightarrow (X \rightarrow X)$$ $$\tilde f_n (\lambda) = R(n \cdot 2 \pi)$$ Now $\tilde f_1$ is an isotopy which parametrises the homeomorphisms from the identity back to itself via just one rotation. Whereas $\tilde f_2$ goes around twice. Clearly there is a loop here, so the space of homeomorphisms is not simply connected.

This hypothesis is confirmed by your other source: Theorem 1.14 (page 43) of A Primer on Mapping Class Groups (Benson Farb, Dan Margalit) which says that we require at least 2 punctures in the plane (equivalently, 3 punctures in the sphere $S^2$) for the space of homeomorphisms isotopic to the identity ($\mathrm{Homeo}_0(S)$ where $S$ is our space) to be contractible.


Let $\tilde f$ be an isotopy from the identity to itself. $$\tilde f : [0,1] \rightarrow (X \rightarrow X)$$ I don't have the ability to show this rigorously, but it seems sensible to me to say that the homotopy of the path of any point $x$ under this isotopy $$\tilde f(\lambda,x) : [0,1] \rightarrow X$$ will be the same. Unfortunately, this seems to be the crux of the question.

Supposing it is true: if you choose isotopy loops $\tilde f$ and $\tilde g$ (from the identity to itself) with the same homotopy in $\mathrm{Homeo}(S)$, then you will find that $\alpha$ and $\beta$ share the same homotopy too. For a sphere with at least 3 holes punched into it, there is only the identity homotopy in the component of $\mathrm{Homeo}(S)$ containing the identity (denoted $\mathrm{Homeo}_0(S)$), and so this is automatically guaranteed. This guarantee cannot be made when $\mathrm{Homeo}(S)$ fails to be contractible, as is the case for the annulus.

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