- #1

- 1

- 0

^{m}-> S

^{p}is homotopic to a constant.

This is the problem 5 in chap8. of 'topology from the differentiable viewpoint(Milnor)'.

I proved it when the map is not onto. But I think it can be onto.

Please help me.

You should upgrade or use an alternative browser.

- Thread starter Stiger
- Start date

- #1

- 1

- 0

This is the problem 5 in chap8. of 'topology from the differentiable viewpoint(Milnor)'.

I proved it when the map is not onto. But I think it can be onto.

Please help me.

- #2

quasar987

Science Advisor

Homework Helper

Gold Member

- 4,783

- 18

It can indeed be onto. For instance, for M=S^1 and S^p=S², take f:S^1-->[0,1]² one of the infamous space-filling curve (loop) (http://en.wikipedia.org/wiki/Space-filling_curve). Then make [0,1]² into S² by identifying all the edges together. Then p o f:S^1-->S² is a continuous surjection, where p:[0,1]²-->[0,1]²/~=S² is the quotient map.

- #3

Office_Shredder

Staff Emeritus

Science Advisor

Gold Member

- 4,525

- 571

^{m}-> S^{p}is homotopic to a constant.

This is the problem 5 in chap8. of 'topology from the differentiable viewpoint(Milnor)'.

I proved it when the map is not onto. But I think it can be onto.

Please help me.

Is the map maybe supposed to be differentiable?

- #4

- 489

- 0

Maybe all you need is that it's homotopic to a smooth map :)

(Then Sard's Theorem)

Share:

- Replies
- 9

- Views
- 2K

- Replies
- 3

- Views
- 4K

- Replies
- 2

- Views
- 2K