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Feb 3, 2014 at 17:28 comment added Don @Will Sawin : The approach via Grayson's flow is lovely, but I was hoping for something more topological (say, using techniques like those in Gramain's paper that Ebert mentions in a comment).
Feb 3, 2014 at 16:28 comment added Will Sawin @Don: Is Sam's argument purely topological?
Feb 3, 2014 at 4:47 comment added Don @Will Sawin : this kernel is Z, and is generated by the mapping class associated to the loop in question. This is Theorem 3.18 in Farb-Margalit's primer. You've actually discovered my secret motivation, which is to give a purely topological proof of this (running your argument backwards!)
Feb 3, 2014 at 4:16 comment added Will Sawin The $S^1$ is the $SO_2$ inside $SO_3$, I believe. With regards to your second question, one would have to compute the kernel of the homomorphism from the mapping class group of the surface cut along the circle to the mapping class group of the surface.
Feb 2, 2014 at 23:12 comment added Sam Nead Slick. I think there may be a missing factor of $S^1$ someplace, as $X$ is the space of parametrized embedded loops? Anyway - can we use the fact that the components of $\operatorname{Homeo}(S)$ are contractible (for generic $S$) to answer the original question this way?
Feb 2, 2014 at 21:56 history answered Will Sawin CC BY-SA 3.0