Compute $\pi^n(S^1\times S^{n+1})$.

Question

What is the space of homotopy classes of maps $S^1\times S^{n+1}\to S^n$? Is there a simple way to compute it, if we know $[S^{n+1}, S^n]\simeq\mathbb{Z}^2$ (resp. $\mathbb{Z}$ for $n=2$)?

Answer 1

Here's an attempt assuming you are interested in unbased homotopy classes of maps.

Let $[X,Y]$ denote based homotopy classes of maps, then what we are looking for is the space $[S^1_+ \wedge S_+^{n+1},S^{n}]$ \begin{align} [S^1_+ \wedge S_+^{n+1},S^{n}] &= [S^{n+1}_+,Maps(S^1_+,S^n)] \end{align}

$Maps(S^1_+,S^n)$ is the free loop space $LS^n$.

We have a split fibration $\Omega S^n \rightarrow LS^n \rightarrow S^n$, the splitting is via inclusion of $S^n$ in $LS^n$ as constant loops.

When $n>2$, $LS^n$ is simply connected and we have \begin{align*} [S^{n+1}_+,LS^n] &= [S^{n+1},LS^n]\\ &=\pi_{n+1}(LS^n)\\ &=\pi_{n+1}(\Omega S^n) \oplus \pi_{n+1}(S^n)\\ &=\pi_{n+2}(S^n) \oplus \pi_{n+1}(S^n) \end{align*}

For $n=1$ we have $$[S^2_+,LS^1] = [S^2_+,S^1 \times \mathbb{Z}] = \mathbb{Z} \oplus \mathbb{Z} $$ (see https://mathoverflow.net/a/149664/29548)

Not sure what happens for $n=2$.