# A continuous map $M \to \mathbb{S}^n$ is homotopic to a constant if $dim(M) < n$?

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?

• A continuous such map could certainly be surjective, but every continuous map between smooth manifolds is homotopic to a smooth one. Feb 3, 2021 at 14:10

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: