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There is a canonical map $\pi:S^3\rightarrow S^2$ which is the usual projection map of $S^3$ understood as a fiber bundle over a $S^2$ base space. If we have a general continuous differentiable map $f:S^3\rightarrow S^2$, is it always possible to find a continuous differentiable map $g: S^3\rightarrow S^3$ such that $f=\pi \circ g$?

I don't expect that this is true, but is there a specific example of an $f$ for which this can be shown to fail?

octonion
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    Hint: Suppose for some $p\in S^2$, that $f^{-1}(p)$ is a disjoint union of two copies of $S^1$. Show there is no $g$ in this case. Can you construct such an $f$? – Jason DeVito - on hiatus Aug 18 '21 at 11:56
  • @JasonDeVito, Thanks for your hint, would you mind clarifying a bit? If $g$ is the identity map then for $f=\pi$, $\pi^{-1}(p)$ has one copy of $S^1$, which would be a great circle of $S^3$. I can certainly chose a $g$ that covers $S^3$ twice, and then this would lead to $g^{-1}\circ\pi^{-1}(p)$ having two copies of $S^1$. Does 'disjoint union' mean the copies of $S^1$ are linking each other? If so, I am not sure how to prove it's not possible, but if they are not linking each other, I think I can explicitly show it is possible. – octonion Aug 18 '21 at 13:31
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    This is what I get for doing math before caffeine. My hint is complete nonsense. In fact, the opposite is true: $g$ always exists. This is a consequence of the fact that $H^2(S^3) = 0$ and https://math.stackexchange.com/questions/72166/obstructions-to-lifting-a-map-for-the-hopf-fibration . Sorry about that! – Jason DeVito - on hiatus Aug 18 '21 at 13:52
  • Which "canonical map" are you asking about? There are many different fiber bundles of $S^3$ over $S^2$, each with its own projection map $S^3 \mapsto S^2$, each with different properties. – Lee Mosher Aug 18 '21 at 14:24
  • @JasonDeVito, No problem, thank you for your new comment and linking that question. That question is along the same lines I was trying to get at with this one. – octonion Aug 18 '21 at 14:29
  • @LeeMosher, I called it "canonical" because I had in mind representing points in $S^3$ by a two complex numbers $z_a, a=0,1$ with $|z_0|^2+|z_1|^2=1$, and then the map to $CP^1\sim S^2$ by identifying the phase. Or we can map directly to a 3 component real unit vector $n^i$, by $n^i= \sum_{ab}z_a^* \sigma^i _{ab}z_b$ with $\sigma$ being Pauli matrices. – octonion Aug 18 '21 at 14:34
  • @Lee: Up to orienting the fibers differently, all fiber bundles $S^3\rightarrow S^2$ are bundle isomorphic to the "usual" Hopf bundle. – Jason DeVito - on hiatus Aug 18 '21 at 15:30
  • What do you mean by "continuous differentiable map"? If it's differentiable, it's continuous. – Xenomorph May 09 '22 at 03:58

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