Demonstrating triviality of images of Hurewicz map Suppose I have some nontrivial element $\rho$ of $\pi_n(S^m)$ where $n>m$, for concreteness, view $S^n$ as $\partial \Delta^{n+1}$ where I take the evident fundamental class in $H_n(S^n )$. Obviously, $\rho$ has to lie in the kernel of the Hurewicz homomorphism since the (reduced) homology of $S^m$ is concentrated in degee $m$.
Is it possible to explicitly describe the chain in $C_{m+1}(S^m)$ that demonstrates the triviality? Is it true that the map $\rho$ is always null cobordant (which I believe is stronger than null homologous)?
 A: Here is a map $F: \Omega_*(S^n) \to \Omega_* \oplus \Omega_{*-n}$ which gives an isomorphism. First we write down the map, then we will check why it is an isomorphism. Call the components $F = (F_1, F_2)$.
First we have $F_1 = \Omega_*(S^n) \to \Omega_*(\text{pt})$ the induced map. That is, forget that you have a map $P \to S^n$ for some probing manifold $P$, and just consider $P$ itself as an element of the bordism group.
Second we have $F_2 = \Omega_*(S^n) \to \Omega_{*-n}$ defines as follows. If $\sigma: P \to S^n$ is your bordism class, pick a regular value of $x$. Then $F_2([\sigma]) = [\sigma^{-1}(x)]$. The bordism class is independent of the regular value chosen and of the bordism class of $\sigma$ itself, both of which are easy to see.
First: this is surjective. To see that it has $([M^k], [N^{k-n}])$ in the image, just consider the map $\sigma: M \sqcup N \times S^n \to S^n$, given on one component by the constant map $\sigma(M) = x$ and on the other component by $\sigma(N, p) = p$. It is evident that in this $F_1([\sigma]) = [M]$, because the second factor is null-bordant; and because any $p \neq x$ is a regular value we find $F_2([\sigma]) = [N]$.
Second: this is injective. This is the claim that you wanted to see explicitly; there are of course algebraic proofs but why not just write down the null-bordism.
So suppose $F_2([\sigma]) = [0]$. I claim that $\sigma$ is bordant to a constant map. Then the fact that $F_1([\sigma]) = 0$ finishes the job --- we have a constant map from a null-bordant manifold, so we may null-bord the manifold and extend that map by the constant map.
Proof: Near a regular value, $\sigma$ looks like $\pi_2: N \times D^n \to D^n$, where $N = \sigma^{-1}(p)$ and $D^k$ is a small disc centered at $p$. Take the manifold $P$ and delete $N \times \text{int}(D^n)$, pasting on $W \times S^n$ in its place, where $W$ bounds $N$. Call this $P'$. There is a natural map $\sigma': P' \to S^n$ given by $\sigma$ on $P' - \text{int}(W) \times S^{n-1}$, and on $W \times S^{n-1}$ our map is $\sigma'(W, p) = p$.
I claim that $\sigma$ and $\sigma'$ are bordant. You can obtain the bordism as $P \times I \cup_{N \times D^n \times \{1\}} W \times D^n$, the same way you normally do with surgery. It is clear how $\widetilde \sigma$ extends across this to restrict to $\sigma$ and $\sigma'$ on the respective boundaries: $\widetilde \sigma$ agrees with $\sigma$ on $P \times I$, and on $W \times D^n$ it is given by $\widetilde \sigma(w, p) = p$.
Thus $\sigma$ is bordant to a null-homotopic map. Null-homotoping it, $\sigma$ is bordant to a constant map. We have constructed the explicit chain as desired.

Try doing this in the case of the Hopf map. You will see that (one instantiation of) the manifold we have constructed is $\Bbb{CP}^2 \setminus D^4$, with the projection to $\Bbb{CP}^1$ being a disc-bundle. This is the usual null-bordism of the Hopf map, and works just as well for the quaternionic Hopf map $S^7 \to \Bbb{HP}^1$. I do not have opinions on the octonionic Hopf map.
