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The mapping class group of the closed unit disk $D_n\subset\mathbb{C}$ with $n$ equally spaced points $P$ on the x-axis (0,1) removed is defined as the set of isotopy classes of self-homeomorphisms $D_n\rightarrow D_n$ which fix the boundary: $\phi\partial D_n=\partial D_n$, and permute the distinguished points: $\phi P=P$.

I am considering a representative $\phi:D_n\rightarrow D_n$ of an element of said mapping class group that induces the identity-automorphism of the fundamental group $\pi_1(D_n,x_0)$, i.e. $\phi_*=\text{id}_{\pi_1(D_n,x_0)}$, where $x_0\in\partial D_n$. I am convinced that $\phi$ has to be homotopic to the identity map $\text{id}_{D_n}$ but am unable to prove it. I know that since $\phi$ is a homeomorphism, $\phi^{-1}_*$ has to be the identity-automorphism as well, but I fail to see how I could use this.

Also, since $\phi$ fixes all loops, for every point $x$ I can choose some loop $\gamma$ on which $x$ lies, and get some homotopy $H_\gamma$ to a loop $\phi\circ\gamma$, on which $\phi(x)$ lies. But I don't see how I could combine those homotopies to get a well defined homotopy from $\text{id}_{D_n}$ to $\phi$, or whether that is even possible at all.

A hint would be appreciated.

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    $\begingroup$ The fundamental group of a disk is trivial, so all automorphisms are the identity. Do you mean the fundamental group of the punctured disk? $\endgroup$ – Cheerful Parsnip Sep 2 '17 at 16:37
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Hint: Choose a basepoint-preserving homotopy equivalence $f:X\to D_n$, where $X$ is a wedge of $n$ circles. Since $f$ is a homotopy equivalence, it suffices to show that $\phi\circ f\simeq id_{D_n}\circ f=f$. To show this, you can show that there is a basepoint-preserving homotopy between $\phi\circ f$ and $f$ when restricted to each of the circles of $X$.

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  • $\begingroup$ I tried this for some time now. I don't know which map you restrict when you say 'when restricted to each of the circles of X'. I also don't understand how I am supposed to construct a homotopy $\phi\circ f\sim f$. I can give paths $\gamma:[0,1]\rightarrow X$ and then argue that $f\circ\gamma\sim\gamma$, but this is something I already knew without introducing X. Could you provide me with more detail on how to construct the homotopy $\phi\circ f\sim f$ please. This has to work by elementary means somehow, I didn't take a course in algebraic topology. I only know the def. of fundamental group. $\endgroup$ – azureai Sep 2 '17 at 20:16
  • $\begingroup$ The space $X$ is just a union of circles $X_1,\dots,X_n$, which overlap at a single point. To construct a homotopy from $\phi\circ f$ to $f$, you can just construct a homotopy from $\phi\circ f|_{X_i}$ to $f|_{X_i}$ for each $i$ (fixing the basepoint) and then glue those homotopies together. To get a homotopy from $\phi\circ f|_{X_i}$ to $f|_{X_i}$, notice that $f|_{X_i}$ is just a map from a circle to $D_n$, and homotopy classes of such maps are the same thing as elements of $\pi_1(D_n)$. $\endgroup$ – Eric Wofsey Sep 2 '17 at 20:32
  • $\begingroup$ So by wedge you mean something like a 'finite hawaiian earring' where all circles are glued at a single point, as opposed to something like this: oooo ? $\endgroup$ – azureai Sep 2 '17 at 20:38
  • $\begingroup$ Yes, an $n$-fold wedge sum of circles. $\endgroup$ – Eric Wofsey Sep 2 '17 at 20:55
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    $\begingroup$ You can just use the pasting lemma, viewing $X\times [0,1]$ as the union of the closed subspaces $X_i\times [0,1]$. $\endgroup$ – Eric Wofsey Sep 3 '17 at 1:50

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