Let $n,q$ be positive integers. We are interested to the cases where $n>q$.

Let $F:mathbb B^ntomathbb S^{q-1}$ be a continuous (differentiable, if needed) map, such that $F(1,0^{n-1})=(1,0^{q-1})$, $F(-1,0^{n-1})=(-1,0^{q-1})$ and $F({0}^1timesmathbb S^{n-2})subset{0}^1timesmathbb S^{q-2}simeqmathbb S^{q-2}$.

Can we find a map $G:mathbb B^{n-1}to mathbb S^{q-2}$ such that $G(theta)=F(0,theta)in mathbb S^{q-2}$ for all $thetainmathbb S^{n-2}$?

MathOverflow Asked by yaoliding on November 14, 2021

1 AnswersThere exists nullhomotopic maps preserving codimension one equators such that the restriction to the equators is not nullhomotopic. Very easy examples come from taking non nullhomotopic maps of spheres and then extending them to a sphere of dimension higher by a nonsurjective map.

However, one can in fact come up with much stranger examples. I claim that for any even n, one can construct a map $S^{2n} rightarrow S^{n+1}$ so that the restriction of the domain to every (2n-1)-sphere $x_1^2 + dots x_{2n} ^2=t$ and codomain to the n-sphere $x_1^2 + dots x_n ^2=t$ is nonnullhomotopic, and this map is nullhomotopic.

This fact follows immediately from the statement, "There is a nonnullhomotopic map $S^{2n-1} rightarrow S^n$ such that the suspension of this map is trivial." This is because the suspension of the map is given by sending meridians to meridians via the map we are suspending.

Such a nullhomotopic map (for $n=2$) exists for the following reason: the Hopf map $S^3 rightarrow S^2$ is a map that has infinite order in the homotopy groups of $S^2$. By a cute argument involving complex conjugation, the suspension of the Hopf map has order 2. Hence, twice the Hopf map is in the kernel of the suspension homomorphism, so we have proven a specific case.

For general n, this follows by computing the rational homotopy groups of $S^n$ and finding only one nontorsion group and applying the above reasoning.

Answered by Connor Malin on November 14, 2021

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