The claim in the title is proved on pp.19-20 of Topological rigidity for non-aspherical manifolds by M. Kreck and W. Lueck. Is there an earlier (classical) reference?
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asked Feb 16, 2021 at 1:19
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1$\begingroup$ I don't know an earlier proof, but you might be interested in the following paper which shows that the mapping class group of $S^1 \times S^2$ is $(\mathbb{Z}/2)^2$: H. Gluck, The embedding of two-spheres in the four-sphere. Trans. Amer. Math. Soc. 104 (1962), 308–333. $\endgroup$– Andy PutmanCommented Feb 16, 2021 at 3:17
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$\begingroup$ @AndyPutman: thanks! Sawashita in theorem 8.8 of [On the Group of Self-Equivalences of the Product of Spheres", Hiroshima Math J, 1975] projecteuclid.org/download/pdf_1/euclid.hmj/1206136785 computed the group of based homotopy self-equivalences of $S^1\times S^2$; it maps onto $(\mathbb Z_2)^2$ with kernel $\mathbb Z$, and presumably, one can compare the two results and extract what I want. I was just hoping for a clean reference. $\endgroup$– Igor BelegradekCommented Feb 16, 2021 at 3:35
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1$\begingroup$ Yes, I think that does it. You could also easily prove from Gluck's theorem that the mapping class group of basepoint-preserving orientation-preserving diffeomorphisms is an extension of $(\mathbb{Z}/2)^2$ by $\mathbb{Z}$ (by the way, I don't know if this split or not). I recently wrote a paper on connect sums of $n$ copies of $S^1 \times S^2$ where we improve a theorem of Laudenbach to show that the mapping class group is a split extension of $Out(F_n)$ by $(\mathbb{Z}/2)^n$, which reduces to Gluck's theorem for $n=1$. See here: arxiv.org/abs/2012.01529 $\endgroup$– Andy PutmanCommented Feb 16, 2021 at 4:06
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1$\begingroup$ @AndyPutman: For $S^1\times S^2$ the mapping class group fixing a basepoint is the same as the one with the basepoint not fixed. This is because in the Birman exact sequence the map $\pi_1 {\rm Diff}(S^1\times S^2)\to\pi_1(S^1\times S^2)$ induced by evaluating diffeomorphisms at the basepoint is surjective. (For others: the Birman exact sequence is the exact sequence of homotopy groups for the fibration ${\rm Diff}(S^1\times S^2)\to S^1\times S^2$ given by evaluating at a basepoint, with fiber the basepoint-preserving diffeomorphisms.) $\endgroup$– Allen HatcherCommented Feb 16, 2021 at 14:55
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1$\begingroup$ @IgorBelegradek: The theorem of Sawashita appears to be incorrect in the case of $S^1\times S^2$. The kernel is ${\mathbb Z}_2$ rather than ${\mathbb Z}$ (and the extension is a product). The ${\mathbb Z}$ presumably comes from $\pi_1SO(2)$ but it should be $\pi_1SO(3)$, corresponding to a homeomorphism of $S^1\times S^2$ rotating $\{t\}\times S^2$ by the angle $t$. As in my comment to Andy Putman there is no difference between basepoint-preserving homotopy equivalences and those that do not preserve basepoint in this situation. $\endgroup$– Allen HatcherCommented Feb 16, 2021 at 15:36
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My guess is that the oldest reference might be Pontryagin's 1941 paper on the homotopy classification of maps from a 3-dimensional complex to the 2-sphere, the English version of which is in Recueil Mathématique 51 pp. 331-359. The application to homotopy equivalences of $S^1\times S^2$ is in an example on page 356 at the end of Section 4 of the paper.
answered Feb 16, 2021 at 5:12
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$\begingroup$ 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. $\endgroup$ Commented Feb 16, 2021 at 14:24
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1$\begingroup$ 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. $\endgroup$ Commented Feb 16, 2021 at 14:26
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1$\begingroup$ 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 $S^1\times S^2$. $\endgroup$ Commented Feb 16, 2021 at 16:52
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2$\begingroup$ 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. $\endgroup$ Commented Feb 16, 2021 at 17:08