0
$\begingroup$

I am reading Farb's book on Mapping Class Groups. For the proof of disc's mapping class group, I have a question. The fact that $Mod(D^2)$ is trivial (for Mod, the definition is orientation preserving homeomorphisms being identity on the boundary up to isotopy) is proved as follows. Say $\phi:D^2 \longrightarrow D^2 $ so that $\phi_{\partial A}$=identity then

$$F(x,t)=\begin{cases} (1-t)\phi(\dfrac{x}{1-t}) & 0\leq|x|< 1-t \\ x & 1-t\leq |x| \leq 1 \end{cases} $$ for $0\leq t<1$ and defining F(x,1) to be the identity map of $D^2$ result in an isotopy F from $\phi$ to identity.

Now my question is, since $D^2$ is convex, every map on $D^2$ is homotopic. I am aware that this does not suffice since we consider isotopies. But before this proof there is the result, saying that every homotopy between orientation preserving homeomorphisms are always isotopies (i.e at all times they are homeomorphisms). So the homotopy that I have in mind becomes isotopy by the result, moreover since $\phi_{\partial D^2}$=identity we have that the isotopy at all times will ve identity on the boundary.

So probably there is a mistake argument, since this one is the first thing that comes to min but not used. Can you help me with what is wrong?

$\endgroup$
1
$\begingroup$

It is not true that "every homotopy between orientation preserving homeomorphisms" of $D^2$ is an isotopy.

It is very easy to write down a counterexample. Take a non-injective continuous map $f : D^2 \to D^2$ which is the identity on the boundary. Use convexity, as you state, to construct a homotopy between the identity and $f$. Now run that homotopy forward from time $0$ to time $1/2$, and backward from time $1/2$ to time $1$, and you obtain a homotopy from the identity map to itself which is not an isotopy.

Alternatively, one can do the construction by application of the homotopy extension lemma, which has the advantage that it extends to any connected surface. Here's how that construction looks like on $D^2$. Pick a point $p \ne 0 \in D^2 - \partial D^2$. From time $t=0$ to time $t=1/2$, homotope $p$ to $0$, let $0$ be stationary, let the points of $\partial D^2$ be stationary, and then apply the homotopy extension lemma to obtain a homotopy from the identity map at time $0$ to a non-injective map at time $1/2$. Now run that homotopy backwards from time $1/2$ to time $1$. Concatenating these, one obtains a homotopy from the identity map to itself which is not an isotopy.

What is true is that for any two orientation preserving homeomorphisms of $D^2$, there exists a homotopy between them which is an isotopy.

$\endgroup$
  • $\begingroup$ I got this,clarification was helpful; yet I think I can correct it. There is this result saying that for arbitrary surface M if there are two orientation preserving homeomorphisms homotopic relative to boundary, then they are also isotopic relative to boundary. So after saying that $\phi$ and identity are homotopic relative to boundary with straight line homotopy, can I conclude they are also isotopic with respect to boundary so that every element in Mod(D) is identity? $\endgroup$ – Lucky Nov 5 '17 at 16:02
  • 1
    $\begingroup$ Yes, that is correct. $\endgroup$ – Lee Mosher Nov 5 '17 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.