Surjection from cohomotopy to cohomology Given a finite topological simplicial complex $X$ of dimension $m$. There exists a canonical way to define a homomorphism from the n-th cohomotopy group of $X$ to the n-th cohomology group.
I would like to know when this homomorphism is surjective?
I only find in the literature some answers for special cases (thanks to Hopf's Theorem on Maps to Spheres):

*

*It is an isomorphism if $n=m$.

*It is an isomorphism if $n=1$.

Are the other general cases studied? Otherwise, was it hard to study them? Or is the answer so obvious?
Thank you
Best
 A: Identify
$$\pi^nX=[X,S^n]\quad \text{and}\quad H^nX=[X,K(\mathbb{Z},n)].$$
The comparison map
$$\Phi:\pi^nX\rightarrow H^nX$$
is induced by a choice of homotopy class of map $\varphi:S^n\rightarrow K(\mathbb{Z},n)$ generating $\pi_nK(\mathbb{Z},n)\cong\mathbb{Z}$. That is
$$\Phi(\alpha)=\varphi\circ\alpha.$$
In general, $\Phi$ is a homomorphism only when either, $X$ is a co-H-space, $S^n$ is an H-space (ie. $n=0,1,3,7$), or when $\dim X=m<2n-2$.
Now, $\Phi$ being surjective is equivalent to the problem of whether for every map $\alpha:X\rightarrow K(\mathbb{Z},n)$ there is a map $\widetilde\alpha:X\rightarrow S^n$ such that $\alpha\simeq\varphi\circ\widetilde\alpha$. e.g. the following diagram commutes up to homotopy
$$\require{AMScd}\begin{CD} X @>\widetilde\alpha>> S^n \\
@| @VV\varphi V \\
X @>\alpha>> K(\mathbb{Z},n)\end{CD}.$$
Problems like this can be solved using obstruction theory. Namely, there is an iteratively defined sequence of obstructions
$$\mathfrak{o}_k(\alpha)\in H^{n+k+1}(X;\pi_{n+k}S^n),\qquad k\geq1.$$
In general, $\mathfrak{o}_k(\alpha)$ depends on choices made in defining each of $\mathfrak{o}_1(\alpha),\dots,\mathfrak{o}_{k-1}(\alpha)$. However, the existence of a vanishing sequence is necessary and sufficient condition for the existence of $\widetilde\alpha$. A good introduction to obstruction theory is found in Davis and Kirk's book Lecture Notes in Algebraic Topology (see especially $\S7.10$. Note that the problem being homotopical we may replace $\varphi$ with a homotopically equivalent fibration).
Now, for an example where the obstructions do not vanish, consider the following. With $X=\mathbb{C}P^m$ for $m\geq2$ we have $H^2\mathbb{C}P^m\cong\mathbb{Z}$ with generator $u$. In case $n=2$ we have $\pi_3S^2\cong\mathbb{Z}$ and $H^4(\mathbb{C}P^m,\pi_3S^2)\cong\mathbb{Z}$. It is not difficult to compute the first obstruction, finding that it is the nontrivial class
$$\mathfrak{o}_1(u)=u^2\in H^4(\mathbb{C}P^m;\mathbb{Z}).$$
In particular $u\in H^2(\mathbb{C}P^m)$ is not in the image of $\Phi$. In fact, no nonzero multiple of $u$ is in the image of $\Phi$ and indeed $\Phi:\pi^2\mathbb{C}P^2\rightarrow H^2\mathbb{C}P^2$ is the zero function.
This example can be pushed further. We can compute $\pi^2\mathbb{C}P^2=0$ and by induction show that $\pi^k\mathbb{C}P^m$ is finite for each finite $m\geq2$. Thus
$$\Phi:\pi^2\mathbb{C}P^m\rightarrow H^2\mathbb{C}P^m$$
is not surjective for any $m\geq2$.
For an example in which $\Phi$ is a homomorphism, take $X=\Sigma\mathbb{C}P^3$ and $n=3$. Since $S^3\cong SU(2)$ has a multiplication, $\pi^3(\Sigma\mathbb{C}P^3)$ is a group, which is abelian by an Eckmann-Hilton argument. Both $\pi^2(\Sigma\mathbb{C}P^2)$ and $H^3(\Sigma \mathbb{C}P^3)$ are isomorphic to the integers, but the image of
$$\Phi:\pi^3(\Sigma\mathbb{C}P^2)\rightarrow H^3(\Sigma\mathbb{C}P^3)$$
has index $2$. To see this take $\alpha\in H^3(\Sigma\mathbb{C}P^3)$ and compute the primary obstruction $\mathcal{o}_1(\alpha)=Sq^2\alpha$, which vanishes iff $\alpha$ is divisible by $2$.
