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Are there continuous functions $f:I\to S^2$ such that $f^{-1}(\{x\})$ is infinite for every $x\in S^2$?

Here, $I=[0,1]$ and $S^2$ is the unit sphere.

I have no idea how to do this.

Note: This is not homework! The question came up when I was thinking about something else.

3SAT
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Akiva Weinberger
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    Could the downvoter please explain what motivated him/her to downvote this question? – Stefan Hamcke Jan 03 '16 at 20:05
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    How about the composition of (i) a space-filling curve from $[0,1]$ to the unit ball, and (ii) a projection from the unit ball to its boundary $S^2$? –  Jan 03 '16 at 20:09
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    The reason this came up is, I was wondering why my book had such a complicated proof that the sphere is simply connected (every loop can be retracted to a point). Then I realized, maybe it's hard to prove because some loops are very complicated. – Akiva Weinberger Jan 03 '16 at 20:10
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    @Rahul You will have problems with mapping the ball's center, unless you have an idea on how to avoid that. – Wojowu Jan 03 '16 at 20:11
  • @AkivaWeinberger: Your realization is spot on. The complicated first step of the proof of simple connectivity is designed to reduce to the case that $f : I \to S^2$ is not surjective, by constructing a path homotopy from an arbitrary $f$ to a nonsurjective $f$. – Lee Mosher Jan 03 '16 at 21:37
  • @LeeMosher The proof I know basically partitions the loop into finitely many nonsurjective pieces (Algebraic Topology by Hatcher), but I'm guessing it's equivalent to the proof you're talking about. – Akiva Weinberger Jan 03 '16 at 21:40

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Consider a space filling curve $\gamma: I \rightarrow I^2$, the projection $q: I^2 \rightarrow S^2$ given by the quotient topology on the square that furnishes the sphere, and the projection $\pi: I^2 \rightarrow I$ on the first coordinate.

The map $q \circ \gamma \circ \pi \circ \gamma$ satisfies what you want.

Aloizio Macedo
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  • So $\pi\circ\gamma$ is an infinity-to-one function from $I\to I$, and $q\circ\gamma$ is a surjective function from $I$ to $S^2$? – Akiva Weinberger Jan 03 '16 at 20:14
  • @AkivaWeiberger Yes. Maybe the usual construction of a space-filling curve already gives a infinity-to-one function (I believe this is the case), but since I don't remember the details well enough, I prefered to make this workaround. – Aloizio Macedo Jan 03 '16 at 20:16
  • I doubt the usual ones (ex: Hilbert and Peano curves) are infinity-to-one, though I encourage you to post it as a question. – Akiva Weinberger Jan 03 '16 at 20:24
  • Check out this related question: Countable-infinity-to-one function – Akiva Weinberger Jan 04 '16 at 18:38
  • If you use the usual construction, shown in many books, for a continuous surjection $g : I\to I^2$ then the composite $p= q g,$ where $q$ is projection to first co-ordinate, then for any nbhd $B$ of any $t\in I$, the cardinal of $ {s\in B :p(s)=p(t)}$ is the cardinal of the reals. And $p$ is nowhere differentiable – DanielWainfleet Jan 04 '16 at 23:12