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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.

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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
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
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
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").

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