Hopf's theorem on cohomotopy group

Question

It is known that if $X$ is a CW complex of dimension at most $p$, then the set $[X, S^{p}]$ has an abelian group structure, which is isomorphic to $H^{p}(X;\mathbb{Z})$. Although I don't know the proof of this theorem, it seems quite intuitive. I want to know what happens when the dimension of $X$ is larger than $p$. If we impose more conditions on $X$ except $\dim X \leq p$, can we give a natural (abelian) group structure on $[X, S^{p}]$? If it is, what is a relation between $[X, S^{p}]$ and $H^{p}(X;\mathbb{Z})$? How far are they? Thanks in advance.

Answer 1

The general theorem here is that for any CW-complex $X$ and abelian group $G$, $[X, K(G, n)]$ has an abelian group structure and is naturally isomorphic to $H^n(X; G)$. The fact you mention follows because if $G = \Bbb Z$, $K(\Bbb Z, n)$ has a model whose $(n+1)$-skeleton is $S^n$ (since $\pi_i K(\Bbb Z, n) = \pi_i S^n$ for all $0 \leq i \leq n$, you only need to start attaching cells to $S^n$ from dimension $n+2$ and beyond). So if $X$ is an $n$-dimensional complex, by cellular approximation theorem $[X, K(\Bbb Z, n)] = [X, S^n]$.

The group structure on $[X, K(G, n)]$ can be read off as follows. If $G$ is abelian, the multiplication map $G \times G \to G$ is a group homomorphism, so induces a multiplication map at the level of spaces $K(G, n) \times K(G, n) \to K(G \times G, n) \to K(G, n)$. This makes $K(G, n)$ into a homotopy-associative, homotopy-commutative $H$-space with a homotopy-identity and having homotopy-inverses. Dualizing gives an abelian group structure on $[X, K(G, n)]$.

To give a sketch of an idea of where the isomorphism comes from, note that $H_n(K(G, n); G) = G$ by either Hurewicz and then homology universal coefficient theorem, or explicitly constructing a model for $K(G, n)$ whose $(n+1)$-skeleton is the Moore space $M(G, n)$. Then by the universal coefficient theorem $H^n(K(G, n); G) = \text{Hom}_\Bbb Z(G, G)$, so let $\alpha$ be the class corresponding to the identity homomorphism $G \to G$. The candidate for the isomorphism $[X, K(G, n)] \to H^n(X; G)$ is given by $[f] \mapsto H^n(f)(\alpha)$. One then verifies that $[-, K(G, n)]$ is an ordinary cohomology theory in the sense of Eilenberg-Steenrod (this is the crux of the theorem and the hard part) and the above map is a natural transformation between the two ordinary cohomology theories which is an isomorphism on $X = pt$, therefore is an isomorphism of cohomology theories.

In general $[X, S^n]$ is very far from $H^n(X; \Bbb Z)$. Consider $X = S^m$ for $m > n$; then $[S^m, S^n] = \pi_m S^n$ are homotopy groups of spheres, which are absolutely horrible things, whereas $H^n(S^m; \Bbb Z)$ is $\Bbb Z$ if $m = n$ and zero otherwise. Note that if $n = 1$, however, since $K(\Bbb Z, 1) = S^1$, $[X, S^1]$ is isomorphic to $H^1(X; \Bbb Z)$ for all $X$ as abelian groups without any dimension restriction on $X$.

I'd like to sketch an argument to show why $[X, S^1]$ is the only cohomotopy group which has a natural abelian group structure. Suppose $[X, S^n]$ has a natural product structure, that is to say, a map $[X, S^n] \times [X, S^n] \to [X, S^n]$ such that the class of the constant map $X \to S^n$ constitutes an (both-sided) identity in a way that is natural with respect to maps $X \to Y$. Consider the map $[X, S^n \times S^n] \to [X, S^n] \times [X, S^n]$ sending the class of $f = (f_1, f_2) : X \to S^n \times S^n$ to $([f_1], [f_2])$. This is natural with respect to maps as well, so by composing we get a map $[X, S^n \times S^n] \to [X, S^n]$ which is a natural transformation of representable functors $\text{Top} \to \text{Set}$, therefore by Yoneda lemma gives a map $S^n \times S^n \to S^n$ which I believe makes $S^n$ into an $H$-space. By Adam's theorem this is possible only if $n = 1, 3, 7$ and only in the case of $n = 1$ we admit a commutative $H$-space structure. So I think this means $[X, S^n]$ does not admit a natural abelian group structure for $n > 1$. It seems interesting to ask what the group structure on $[X, S^3]$ is, where $S^3 = SU(2)$ is equipped with quaternionic multiplication. For a rather simple observation, $[X, SU(2)] = [\Sigma X, BSU(2)]$ and $BSU(2) = \Bbb{HP}^\infty$ so $[X, SU(2)]$ classifies quaternionic line bundles on $\Sigma X$ by the clutching construction so $[X, S^3]$ is isomorphic to the group of quaternionic line bundles on $\Sigma X$ where the product is given by tensor product over $\Bbb H$ (this is noncommutative, simply because quaternionic multiplication is noncommutative). I don't know if there's a simpler description of this "quaternionic Picard group" though.