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Wednesday, July 4, 2018

About the Retractions of a surface of genus $g$ onto circles.

Exercise. In the surface $M_g$ of genus $g$, let $C$ be a circle that separates $M_g$ into two compact subsurfaces $M_h', M_k'$ obtained from the surfaces $M_h,M_k$ by deleting an open disk from each. Show that $M_{h'}$ doesn't retract onto its boundary circle $C$, and hence $M_g$ doesn't retract onto $C$.
Solution.  When we deal with $M_g$, are dealing with cell complexes like the following:

built by pasting $2g$ $1-$cells $\alpha_1,\beta_1,\alpha_2,\beta_2,\dots, \alpha_g,\beta_g$ to a $0-$cell, and then pasting a $2-$cell by the map $S^1\to X^1$ given by $[\alpha_1,\beta_1][\alpha_2,\beta_2]\cdots [\alpha_g, \beta_g]$.
When a disk is removed, the resulting surface $M_g'$
deform-retracts to the $1-$skeleton
whose fundamental group is $\Large{*}_{\normalsize i=1}^{\normalsize2g}\normalsize\bZ$. Now, let's suppose that $M_h'$ retracts onto $C$. Then, the homomorphism $i_*:\bZ\to \Large{*}_{\normalsize i=1}^{\normalsize2h}\normalsize\bZ$ is injective. We can then, quotient everything over its commutator (Since $i_*$ is injective) to get a homomorphism $i^*:\bZ\to \bigoplus_{i=1}^{2h} \bZ$ which, since $\bigoplus_{i=1}^{2h} \bZ$ has no torsion, is also injective. This is a contradiction, since $i_*(1)=[e_1,e_2][e_3,e_4]\cdots [e_{2h-1},e_{2h}]\mapsto 0$ when passing to the quotient.
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