Prove $S_g$ retracts to $C$

Question

Let $S_g$ be the closed oriented surface of genus g. Let $C$ be a simple closed curve in $S_g$. Prove that $S_g$ retracts to $C$ if and only if $C$ does not separate $S_g$.

If $S_g$ retracts to $C$, by Alexander-Lefschetz duality, I can get $H_0(S_g-C)=\mathbb{Z}$, therefore $C$ does not separate $S_g$. But I have no idea how to prove another direction.

score 1 · Answer 1 · answered Jan 03 '14 at 03:42

1

I don't think the statement is true. Take $S_g$ with $g>1$, take $C$ around one of the holes, and $S_g$ won't retract to $C$ because such rectraction would imply that $S_{g-1}$ retracts to a point.

answered Jan 03 '14 at 03:42

Michael

1,726

Doesn't every space retract to a point via the constant map? – Math 2tor Nov 08 '23 at 17:50

score 1 · Answer 2 · answered Jan 03 '14 at 03:46

1

Cut along the curve to get an annulus with handles Quotient each handle to a point and you get just an annulus. Map this to the circle by the natural retraction. This all descends to a continuous map on the original space, since each point on the other copy of $C$ (the top of the annulus) gets mapped to the corresponding point in the other copy of $C$.

answered Jan 03 '14 at 03:46

Brian Rushton

13,375

How is quotienting the handles a retraction? – Igor Rivin Jan 03 '14 at 03:55
Hello Igor! It's not a retraction until the quotient point gets mapped to the boundary circle. – Brian Rushton Jan 03 '14 at 03:57
Well, it kind of is. The point is that a retraction is just a continuous map which is the identity on $C$ (in this case), and the collapsing of the handles is a continuous map. I think a lot of people confuse a retraction with a deformation retraction. – Igor Rivin Jan 03 '14 at 04:00
I frequently confuse the two. I also think that all homeomorphic subsets of 3-space are isotopic sometimes. – Brian Rushton Jan 03 '14 at 04:05
I don't believe you. You are a student of Jim Cannon, so wise in the ways of topology. – Igor Rivin Jan 03 '14 at 04:11
It sure makes knot theory easier. – Brian Rushton Jan 03 '14 at 04:12
It's almost too easy that way. – Igor Rivin Jan 03 '14 at 04:13

Prove $S_g$ retracts to $C$

2 Answers2