# Infinity-to-one function

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.

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Could the downvoter please explain what motivated him/her to downvote this question? – Stefan Hamcke Jan 3 at 20:05
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 $S^2$? – Rahul Jan 3 at 20:09
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 3 at 20:10
@Rahul You will have problems with mapping the ball's center, unless you have an idea on how to avoid that. – Wojowu Jan 3 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 3 at 21:37

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.
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 3 at 20:14
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 – user254665 Jan 4 at 23:12