# The group structure on $[X,S^n]$ induced by the framed bordism

I'm concerned about the group structure on $$[X,S^n]$$, i.e. the set of homotopy classes of continuous maps from $$X$$ to $$S^n$$.

On the one hand, $$[X,Y]$$ has a group structure that is natural with respect to $$X$$ if and only if $$Y$$ is an H-space. The naturality is in the sense that $$f:X'\to X$$ induces a homomorphism $$f_{\star}:[X,Y]\to[X',Y]$$. It's known that $$S^n$$ is an H-space only for $$n=0,1,3,7$$. Thus, for a general $$X$$, there's no natural group structure on $$[X,S^n]$$ when $$n$$ takes other values.

On the other hand, when $$X$$ is a closed smooth $$d$$-manifold, the Pontryagin-Thom construction establishes a bijection between $$[X,S^n]$$ and $$\Omega_{d-n}^{fr}(X)$$, i.e. the (unstable) $$(d-n)$$th framed bordism set of $$X$$. When $$d<2n-1$$, two transverse $$(d-n)$$-submanifolds in $$X$$ have no intersection and the disjoint union induces a group structure on $$\Omega_{d-n}^{fr}(X)$$. Then we can make $$[X,S^n]$$ into a group via the Pontryagin-Thom bijection.

Hence $$[X,S^n]$$ is a group when $$X$$ is a closed smooth manifold whose dimension $$<2n-1$$. I would like to know how "natural" or how "unnatural" this group structure is: (1) if $$X'$$ is a closed smooth manifold whose dimension $$<2n-1$$, does a smooth map $$f:X'\to X$$ induce a group homomorphism? (2) When $$n=0,1,3,7$$, does this group structure coincide with the H-space induced group structure?

Edit 1 The question has been edited to correct a mistake pointed out by Gregory.

Edit 2 As mentioned by Tyrone, $$[X,S^n]$$ is a cohomotopy group when $$X$$ is a complex of dimension $$<2n-1$$, with the multiplication induced by the folding map. It reduces to the H-space induced group if $$n=0,1,3,7$$. My Q2 essentially asks whether cohomotopy groups coincide with framed bordism groups (when the latter applies).

• As a matter of fact, if $d=2n-1$ then it is true that any two transverse $n-1$ submanifolds of a $d$-dimensional manifold are disjoint, but the cobordism class of the disjoint union is not uniquely determined by the cobordism classes of the components. So the disjoint union induces a well-defined group structure on $\Omega^{fr}_{d-n}(M^d)$ only for $d\le 2n-2$. The answer to Q1 can be understood via the suspension map $S^n\to \Omega^\infty \Sigma^\infty S^n$ and the identification $\Omega^{fr}_{d-n}(X)\cong [X, \Omega^\infty \Sigma^\infty S^n]$. Commented Dec 19, 2021 at 9:21
• @GregoryArone Thank you for pointing out my mistake! And indeed, the naturality is evident by considering $\Omega^{\infty}\Sigma^{\infty}S^n$. I still have a further question: in terms of $[X,\Omega^{\infty}\Sigma^{\infty}S^n]$, how to understand the dimension condition $d<2n-1$? It looks like a group for any $n$.
– Leo
Commented Dec 19, 2021 at 12:21
• It is always a group, but the map $[X, S^n] \to [X, \Omega^\infty \Sigma^\infty X]$ is a bijection only if $X$ is a complex of dimension $< 2n-1$. Commented Dec 19, 2021 at 12:25
• @GregoryArone I see. Thank you!
– Leo
Commented Dec 19, 2021 at 12:28
• There's a nice chapter on cohomotopy groups in Hu's book: Sze-tsen Hu, Homotopy theory, Pure and Applied Mathematics, Vol. VIII Commented Dec 22, 2021 at 8:29

This is an answer to question (2). Let $$n=0,1,3,7$$ and $$i=1,2$$ and $$d\leq 2n-2$$. Let $$e=(1,0,\ldots,0)\in S^n$$

Let $$f_i:X\rightarrow S^n$$ be two maps representing framed submanifolds $$(M_i,\nu_i)$$. Let $$T_i$$ be tubular neighborhoods of $$M_i$$.

By the assumptions on the dimensions, the maps can be chosen in such a way that

• $$T_1\cap T_2=\emptyset$$.
• The value $$-e\in S^n$$ is regular for both $$f_1,f_2$$
• $$f_i^{-1}(\{-e\})=M_i$$ and the framing induced by the differential is $$\nu_i$$.
• $$f_i(X\setminus T_i)=e$$.

Let $$g:X\rightarrow S^n$$ be the product map $$g(x)=f_1(x)f_2(x)$$. By the choices made above $$-e$$ is a regular value and $$g^{-1}(\{-e\})=M_1\cup M_2$$. The induced framing on $$M_1\cup M_2$$ is $$\nu_1,\nu_2$$ on the respective components. This shows that the group structures coincide.

• 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. Commented Dec 19, 2021 at 10:43
• 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. Commented Dec 19, 2021 at 10:45
• 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.
– Leo
Commented Dec 20, 2021 at 4:04