Given the Hopf map $h:S^3\to S^2$ and an inclusion $i:S^2\hookrightarrow S^3$, the map $h\circ i:S^2\to S^2$ has mapping degree zero. Therefore, it is homotopic to the constant map and the image of the equator $S^2\subset S^3$ can be contracted to a point. This gives a representative of the sum $[f]+[g]\in\pi\_3(S^2)$ for some $[f],[g]\in\pi\_3(S^2)$. Since the Hopf map represents the class $[h]=1\in\mathbb{Z}=\pi\_3(S^2)$, this would imply that $[f]+[g]=1$, i.e. that the Hopf invariants of $f$ and $g$ sum to $1$. However, it seems awkward that the contraction of the equator of such a canonical map as the Hopf map would yield two maps with Hopf invariant not having the same absolute value. Or is this in fact the case?



Splitting the Hopf map in two
Here is an expanded version of my comment above.


Let $h: S^3\to S^2$ be the Hopf map (or any other pointed map for that matter). Choose a pointed null-homotopy $i\_t: S^2\to S^3$ of the standard inclusion $i:S^2\hookrightarrow S^3$.


The composition $h\circ i\_t$ is a null-homotopy of $h\circ i$, and therefore gives a map $$h': S^3\cup\_i C(S^2) \to S^2$$ defined on the mapping cone of $i$ which extends $h$. We can identify $h'$ up to homotopy with a map $h'=(f\vee g): S^3\vee S^3\to S^2$ satisfying $h'\circ \nu\simeq h$, where $\nu:S^3\to S^3\vee S^3$ denotes the pinch map.


Thus $[h]=[f]+[g]\in \pi\_3(S^2)$, as you say, but it is easy to see that $[f]$ and $[g]$ are going to depend on $i\_t$. I'll leave the details of what goes where as an exercise, but it seems to me that a standard $i\_t$ would result in one or other of $[f]$ and $[g]$ being $0$ and the other being $[h]$. 


