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-   -   retraction in surface of genus g (http://www.physicsforums.com/showthread.php?t=510720)

ForMyThunder Jun29-11 05:25 PM

retraction in surface of genus g
 
1. The problem statement, all variables and given/known data
In the surface Mg of genus g, let C be a circle that separates Mh' and Mk' obtained from the closed surfaces Mh and Mk by deleting an open disk from each. Show that Mh' does not retract onto its boundary circle C, and hence Mg does not retract onto C.

Hatcher Allen. Algebraic Topology Section 1.2 Problem 9


2. Relevant equations



3. The attempt at a solution Suppose there was such a retraction. Then we would have that [tex]i_*:\pi_1(C)\to\pi_1(M'_h)[/tex] induced by the inclusion map is injective and that [tex]\phi:\pi_1(M'_h)\to\pi_1(M_h)[/tex] is surjective with kernel [itex]i_*(\pi_1(C))[/itex]. Thus, [tex]\pi_1(M'_h)/i_*(\pi_1(C))\cong \pi_1(M_h)[/tex] and by taking the abelianizations: [tex]\mathbb{Z}^{2h-1}\cong\mathbb{Z}^{2h}/\mathbb{Z}\cong\mathbb{Z}^{2h}[/tex] yielding a contradiction.

Is this correct? I used the assumption that C was a retract of Mh' to say that the fundamental group of C is isomorphic to a subgroup of the fundamental group of Mh'.

ForMyThunder Jun30-11 05:13 PM

Re: retraction in surface of genus g
 
Actually, in that last line, [itex]\mathbb{Z}^{2h-1}\cong\mathbb{Z}^{2h}/\mathbb{Z}[/itex] is not necessarily true. But the second isomorphism in that line is implied by the previous line and this leads to the contradiction.


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