Retraction in surface of genus g

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SUMMARY

The discussion focuses on the non-retraction of the surface Mh' onto the boundary circle C in the context of algebraic topology, specifically within the surface Mg of genus g. The user demonstrates that assuming a retraction leads to a contradiction involving the fundamental groups, specifically that the isomorphism between \mathbb{Z}^{2h-1} and \mathbb{Z}^{2h}/\mathbb{Z} is not valid. This conclusion confirms that Mh' cannot retract onto C, thereby establishing that Mg does not retract onto C either.

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  • Understanding of algebraic topology concepts, particularly fundamental groups.
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Homework Statement


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


Homework Equations





The Attempt at a Solution

Suppose there was such a retraction. Then we would have that i_*:\pi_1(C)\to\pi_1(M'_h) induced by the inclusion map is injective and that \phi:\pi_1(M'_h)\to\pi_1(M_h) is surjective with kernel i_*(\pi_1(C)). Thus, \pi_1(M'_h)/i_*(\pi_1(C))\cong \pi_1(M_h) and by taking the abelianizations: \mathbb{Z}^{2h-1}\cong\mathbb{Z}^{2h}/\mathbb{Z}\cong\mathbb{Z}^{2h} 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'.
 
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Actually, in that last line, \mathbb{Z}^{2h-1}\cong\mathbb{Z}^{2h}/\mathbb{Z} 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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