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Dec 20, 2021 at 9:28 history edited Thomas Rot CC BY-SA 4.0
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Dec 20, 2021 at 4:04 comment added Leo Thank you and @Tyrone! I skimmed the idea of cohomotopy groups and I think that's what I really want to know. Although $[X,S^n]$ is not a group in general, it is naturally a group for a complex of dimension $<2n-1$. It's wonderful to see that Thomas shows the algebraic-topology idea and the differential-topology idea agree with each other on this issue, as they're supposed to do.
Dec 19, 2021 at 12:50 vote accept Leo
Dec 19, 2021 at 10:45 comment added Tyrone The fact that an H-space structure is exactly an extension of the folding map along the wedge inclusion means that this part comes for free once the details of my previous comment are expanded upon. In any case the naturality is apparent.
Dec 19, 2021 at 10:43 comment added Tyrone The inclusion $S^n\vee S^n\hookrightarrow S^n\times S^n$ is $(2n-1)$-connected. Thus given a complex $X$ of dimension $<2n-1$ and maps $f,g:X\rightarrow S^n$, the map $(f,g):X\rightarrow S^n\times S^n$ has a unique compression into $S^n\vee S^n$. Follow this with the folding map $\nabla:S^n\vee S^n\rightarrow S^n$ to define the product $f+g$. Borsuk shows that this gives a group structure (these are his cohomotopy groups. See "Sur les groupes des classes de transformations continues."). I'm pretty certain that your argument shows that this structure coincides with the bordism group product.
Dec 19, 2021 at 9:55 history edited Thomas Rot CC BY-SA 4.0
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Dec 19, 2021 at 9:49 history answered Thomas Rot CC BY-SA 4.0