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Timeline for extension of surface homeomorphism

Current License: CC BY-SA 3.0

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Jun 12, 2012 at 21:07 history edited Ian Agol CC BY-SA 3.0
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Jun 2, 2012 at 20:22 comment added Sam Nead Arrange matters so that $g$ doesn't send $\lambda^+(f)$ to $\lambda^-(f)$; post-compose $g$ with a separating Dehn twist if necessary. Now use the Brouwer fixed point theorem and the north-south dynamics of $f$ to prove (for $n$ sufficiently large) that the action of $f^n \circ g$ has exactly two fixed points: one close to $\lambda^+(f)$ and one close to $g^{-1}(\lambda^-(f))$.
Jun 2, 2012 at 19:24 comment added Igor Rivin @Sam: why is that?
Jun 2, 2012 at 17:18 comment added Sam Nead Igor - take any pA element $f$ that acts trivially on homology. Take any $g$ with the desired action on homology. Then for some $n$ the element $f^n \circ g$ will be pA.
Jun 2, 2012 at 15:57 comment added Igor Rivin Why does one have a pseudo-ANosov giving any symplectic matrix?
Jun 2, 2012 at 15:11 history answered Ian Agol CC BY-SA 3.0