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I'm trying to solve the following exercise from Milnor's Topology from the Differentiable Viewpoint, but I'm not sure I understood the statement.

Let $M$ be a smooth manifold with $dim(M) = m < n$, show that every map $f : M \to \mathbb{S}^n$ is homotopic to a constant.

I was able to proove this under the assumption that $f$ is a smooth map. In this case, since $m < n$, $df_x$ is not surjective $\forall x\in M$. By Sard's Theorem exists $p_0\in \mathbb{S}^n$ regular value for $f$, so $f^{-1}(p_0)$ must be empty and the map is not surjective and I can build and homotopy between $f$ and the constant $-p_0$ (they are always not opposite).

If $f$ is not smooth, can I still say that it's not surjective? I'm not able to prove that, but I'm not even sure that this is the right way to proceed if $f$ is just continuous. Sometimes in the book the author assumes that maps are smooth, so does this statement still hold for continuous maps?

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    $\begingroup$ A continuous such map could certainly be surjective, but every continuous map between smooth manifolds is homotopic to a smooth one. $\endgroup$
    – Thorgott
    Feb 3, 2021 at 14:10

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The reason is that each continuous $f : M \to S^n$ can be approximated arbitrarily closely by smooth maps. That is, for each $\epsilon > 0$ there exists a smooth map $g : M \to S^n$ such that $\lVert g(x) - f(x) \rVert < \epsilon$ for all $x \in M$. Note that for $\epsilon \le 2$ the maps $g, f$ are homotopic via $$H(x,t) = \frac{tg(x) - (1-t)f(x)}{\lVert tg(x) - (1-t)f(x) \rVert}.$$ Milnor does not prove this, but addresses it in the exercises, problem 4 (at least fort compact $M$). See also his proof of the Brouwer fixed point theorem. Quotation:

The procedure employed here can frequently be applied in more general situations: to prove a proposition about continuous mappings, we first establish the result for smooth mappings and then try to use an approximation theorem to pass to the continuous case. (Compare 58, Problem 4.)

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  • $\begingroup$ Thanks, that's just what I needed. I already solved problem 4 so now it' everything clear for compacts M. I'll check for general M. $\endgroup$
    – IvanB
    Feb 3, 2021 at 14:42

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