# Simply connected slices

Assume $\Omega$ is an open set in $\mathbb R^3$ such that the intersection of $\Omega$ with any horizontal plane is simply connected.

Can you prove that $\Omega$ is simply connected?

(Note that by the definition, simply connected set can not be empty.)

• The proof given by Tom Goodwillie below is done with bare hands. I would prefer to find ready to use tool for answering this and similar questions.
• If you also assume that $\Omega$ is bounded, then each slice is bounded open simply connected, and so its complements is connected. The complements of the slices can be joined, again because of the boundedness, and so $\mathbf{R}^3\backslash\Omega$ is connected. I don't know if this implies simply connected, like on the plane, but perhaps it helps. – erz Jun 24 '17 at 1:58
• If $\Omega$ is bounded then there exists a horizontal plane whose intersection with $\Omega$ is empty and therefore not simply connected. – Timothy Chow Jun 24 '17 at 3:26
• by adding one point at infinity it seems erz's comment can be done without boundedness, but it seems directed rather at showing the absence of non bounding 2 cycles rather than 1 cycles. See the answer below however which seems to exclude both. – roy smith Jun 24 '17 at 15:57
• Does "simply connected" include "connected"? Otherwise there are simple counterexamples. – Alexandre Eremenko Jun 25 '17 at 19:39
• @AntonPetrunin: Actually, there is such a theory. Start from Smale's generalization of Vietoris mapping theorem in [Proc. Amer. Math. Soc. 8 (1957), 604–610.], see maths.ed.ac.uk/~aar/papers/smale3.pdf and trace references from there. Basically the theorem says that under very mild regularity assumptions any proper continuous surjection with $n$-connected fibers induces homotopy groups isomorhism up to dimension $n$, and surjection in dimension $n+1$. Of course you map isn't proper, but maybe by looking around you can find a version that works for you. – Igor Belegradek Jun 26 '17 at 19:58

Yes, I think so. Let's show that every compact set $K\subset \Omega$ is contained in some compact contractible subset of $\Omega$. We use the fact that in a simply connected open subset of the plane every compact set is contained in some compact contractible set.
Denote by $P_t$ the plane $\mathbb R^2\times t$, and define the set $\Omega_t\subset\mathbb R^2$ by $\Omega_t\times t=\Omega\cap P_t$. Define $K_t$ likewise.
For each $t$ choose a compact contractible set $C_t\subset \Omega_t$ such that $K_t\subset C_t$. There must be an interval $J_t$ containing $t$ such that for every $t'\in J_t$ we have $K_{t'}\subset C_t\subset \Omega_{t'}$.
The set of all $t$ such that $K_t$ is nonempty can be covered by finitely such intervals. Thus for some $a$ there are real numbers $s_0<\dots <s_a$ and numbers $t_i\in [s_{i-1},s_i]$ such that $$K\subset \cup_{i=1}^a ([s_{i-1},s_i]\times C_{t_i})\subset \Omega.$$ Enlarge this union to make it contractible by choosing, for each $i=1,\dots a-1$, a compact contractible set $D_i$ such that $C_{t_i}\cup C_{t_{i+1}}\subset D_i\subset \Omega_{t_{i-1}}$ and then adding the sets $s_i\times D_i$.