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some deformation retractions
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FibonacciFan
Riemann Hypothesis
Riemann Hypothesis

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#1
some deformation retractions

1. Construct an explicit deformation retraction of the torus with one point deleted onto a graph consisting of two circles intersecting in a point, namely, longitude and meridian circles of the torus. (Hatcher Exercise 0.1)

2. Construct an explicit deformation retraction of onto . (Hatcher Exercise 0.2)

Attempted Solutions:

1. I can kind of visualize the deformation retraction: draw lines going radially outward from the point removed; all the points move along those lines to either the meridian or the longitude. But I think some of these lines will intersect and I have no idea how to write that down as a formula (I assume that is what explicitly means).

2. I assume the retraction is x/norm(x). As in 1, I think we want to move all the points radially toward S^{n-1}. I don't know how to pace that points so that this will be continuous. Won't the points have to move "infinitely fast" near t = 0?

PostPosted: Mon Aug 18, 2008 7:49 pm  Back to top 
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jmerry
Birch & Swinnerton Dyer
Birch & Swinnerton Dyer

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#2
A deformation retract is a homotopy- which means you have to say how it moves. is a retract, but not a deformation retract.

A way to create a deformation retract: let (or a similar homotopy between and ). Now, let the homotopy .

PostPosted: Mon Aug 18, 2008 8:22 pm  Back to top 
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FibonacciFan
Riemann Hypothesis
Riemann Hypothesis

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#3
jmerry wrote:
A deformation retract is a homotopy- which means you have to say how it moves. is a retract, but not a deformation retract.

A way to create a deformation retract: let (or a similar homotopy between and ). Now, let the homotopy g(t,x) = f(t,|x|)\frac {x}{|x|}.


Yes thanks that solves 2. Intuitively, it seems weird that there could be a homotopy between the identity on and the constant map x to 1 since the range of the homotopy is at all times greater than 0 but at 0 it suddenly becomes finite. Anyway, algebraic operations in are continuous and composition of continuous functions is continuous, so it must be true.

What about 1?

PostPosted: Mon Aug 18, 2008 9:54 pm  Back to top 
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jmerry
Birch & Swinnerton Dyer
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#4
Interpret the torus as a rectangle with opposite sides identified. Cut out a point somewhere in the middle, and then shrink everything away from that point to the rim.

There's no need to quote the entire post immediately preceding yours, ever.

PostPosted: Mon Aug 18, 2008 10:36 pm  Back to top 
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FibonacciFan
Riemann Hypothesis
Riemann Hypothesis

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#5
I see thanks. Can I move this thread to the "Solved Problems" thread or does a moderator need to do that?

PostPosted: Tue Aug 19, 2008 7:53 am  Back to top 
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ZetaX
Birch & Swinnerton Dyer
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#6
\text{Unsolved Problems} \to \frame{\text{some miracle happens}} \to \text{Solved Problems}

PS: No you can't Wink
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PostPosted: Tue Aug 19, 2008 10:49 am  Back to top 
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