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Feb 16, 2021 at 17:22 vote accept Igor Belegradek
Feb 16, 2021 at 17:08 comment added Allen Hatcher Incidentally there is also Laudenbach's 1974 Astérisque volume "Topologie de la dimension trois: homotopie et isotopie" which proves this result for a broad class of 3-manifolds including $S^1\times S^2$, as stated on page 5. The companion question of whether homotopic diffeomorphisms are isotopic is also treated in depth.
Feb 16, 2021 at 16:52 comment added Allen Hatcher You are right that there is no explicit statement here of the result you are looking for, so one has to read between the lines a bit. It does say that there are exactly two non-equivalent mappings $S^1\times S^2 \to S^2$ satisfying the appropriate homology condition, with one mapping obtained from the other by composing with the twist homeomorphism of $S^1\times S^2$, and this seems like the essential point. Surely he would have known what this meant for homotopy equivalences of $S^1\times S^2$.
Feb 16, 2021 at 14:26 comment added Igor Belegradek I don't see where Pontryagin shows that the obstruction is zero. In any case the paper is clearly very relevant- thank you very much.
Feb 16, 2021 at 14:24 comment added Igor Belegradek I don't see that the Example on p.356 of maths.ed.ac.uk/~v1ranick/papers/pont5.pdf makes any specific claim. It seems the strategy is to classify a homotopy self-equivalence $f$ of $M=S^1\times S^2$ by its coordinates, which give automorphisms of $\pi_i(S^i)=\pi_i(M)$, $i=1,2$, where $i=2$ is given by the lift of $f$ to the universal cover. Compositing with self-homeomorphisms we can assume the automorphisms are trivial, ie $f=id$ on $S^1\vee S^2$. The obstruction to homotopy of $f$ to $id$ is in $H^3(M, \pi_3(S^2))=\mathbb Z$, and I guess Pontryagin shows the obstruction is zero.
Feb 16, 2021 at 5:12 history answered Allen Hatcher CC BY-SA 4.0