# For any smooth n-manifold $M$, construct a smooth map $f:M\to S^n$ which is not null-homotopic

PROBLEM: For any smooth n-manifold $M$, construct a smooth map $f:M\to S^n$ which is not null-homotopic ,or even of degree 1

The following is my idea: First, choosing an arbitrary open coordinate ball $B$ on $M$ and then collapsing $M\setminus B$ to a single point gives a continuous map $f:M \to S^n$.Also, we may assume all $M\setminus B$ is mapped to the north pole $\mathbb N$.

It is known that any continuous map $f$ between two manifolds $M_1$ and $M_2$ is homotopic to a smooth map $F$.

Moreover, can we require $|F-f|\le \epsilon$ for any positive continuous function $\epsilon$ on $M_1$ ?

Here the norm $|\cdot|$ can be viewed as a Euclidean norm when we embeds $M_2$ to some $\mathbb R^m$

Personally, I believe it's true. If the statement above is indeed true, then we denote $F$ the smooth map homotopic to $f$ and choose $\epsilon$ small enough. Also, we choose a regular value $\mathbb P$ near the South pole $\mathbb S$ of $S^n$. Then, $\mathbb P$ can only has one preimage and thus the degree of $F$ must be 1 or -1

• This is certainly impossible for $M=\mathbb R^n$ – Georges Elencwajg Jul 2 '15 at 10:48
• I would assume for safety $M$ compact and orientable. – orangeskid Jul 2 '15 at 12:18
• @orangeskid It's false without closed (every map from a non-closed smooth $n$-manifold to $S^n$ is null-homotopic). With closed you don't need orientable - the same standard construction works either way. (To prove it's not null-homotopic you just note it induces an isomorphism on $H_n(M;\mathbb Z_2)$, rather than with $\mathbb Z$ coefficients.) – user98602 Jul 6 '15 at 15:53

Instead of using approximations it's easier to just construct the desired smooth map. Specifically, given a coordinate ball $B\subset M$, we can use a bump function to construct a smooth surjection $B\to S^n$ which is constant on a neighborhood of $\partial B$. This function extends to the entire manifold $M$ by letting $M-B$ map to the same constant, and this map has degree $1$.
• This argument seems to apply equally well if $M=\mathbb{R}^n$, but the map is clearly null-homotopic in that case. – Tad Jul 11 '15 at 13:34
• @Tad This is because a degree one map whose domain is $\mathbb{R}^n$ can be nullhomotopic. (Indeed, every map with domain $\mathbb{R}^n$ is nullhomotopic.) It's only in the case of compact manifolds that a degree one map is necessarily not nullhomotopic. – Jim Belk Jul 11 '15 at 14:17