Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $M$ and $N$ be two topological manifolds. Denote by $[M,N]^{\text{diff}}$ and $[M,N]^{\text{cont}}$ the set of homotopy classes of differential and continuous maps respectively. Is it true that $[M,N]^{\text{diff}}=[M,N]^{\text{cont}}$ ? Any reference? Thank you.

Edit: Thanks to the comments below, I should ask whether it is true when $M$ and $N$ are differentiable manifolds.

share|improve this question
2  
You probably mean "... be two differentiable manifolds ..." for $[M,N]^{diff}$ to make sense. –  Somnath Basu Mar 28 '10 at 3:01
    
I hope that this result is true without any condition on the structure of manifold. In fact, what I am interested is the case in which $M$ is smooth while $N$ may not have smooth structure. –  Fei YE Mar 28 '10 at 3:08
2  
What do you mean by a differentiable map between manifolds, one of which doesn't have a smooth structure? –  Qiaochu Yuan Mar 28 '10 at 4:02
1  
Fei YE, as Yuan and Basu point out, $[M,N]^{diff}$ doens't make sense if $N$ isn't smooth. –  Ryan Budney Mar 28 '10 at 5:56
    
do you require for the homotopies of the maps on the left to be smooth functions themselves? –  Sean Tilson Mar 28 '10 at 22:15

1 Answer 1

up vote 10 down vote accepted

There's no way this can be literally true:

$$[M,N]^{diff} = [M,N]^{cont}$$

Most of the continuous functions from $M$ to $N$ are not differentiable. So there's no way the above equality can be an equality of sets. I think what you want to ask is if the inclusion:

$$[M,N]^{diff} \to [M,N]^{cont}$$

a bijection? This is answered affimatively in Hirsch's "Differential Topology" textbook. It boils down to a smoothing argument, that every continuous function can be uniformly approximated by a $C^\infty$-smooth function and the smoothing is unique up to a small homotopy. The argument goes further, to state the the space of continuous functions has the same homotopy-type as the space of $C^\infty$ functions. The smoothing argument can be done with bump functions and partitions of unity, and also via a standard convolution with a bump function argument ("smoothing operators").

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.