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.
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    $\begingroup$ 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. $\endgroup$
    – erz
    Jun 24, 2017 at 1:58
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    $\begingroup$ If $\Omega$ is bounded then there exists a horizontal plane whose intersection with $\Omega$ is empty and therefore not simply connected. $\endgroup$ Jun 24, 2017 at 3:26
  • $\begingroup$ 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. $\endgroup$
    – roy smith
    Jun 24, 2017 at 15:57
  • $\begingroup$ Does "simply connected" include "connected"? Otherwise there are simple counterexamples. $\endgroup$ Jun 25, 2017 at 19:39
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    $\begingroup$ @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. $\endgroup$ Jun 26, 2017 at 19:58

1 Answer 1


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 $s_i\times D_i$.

  • $\begingroup$ For contrast, note that there is a cycle in the 3x3x3 cubic lattice graph whose three orthogonal projections are simply connected. Thus you need lots of slices to ensure simply connected in general. Gerhard "Two By Two Too Simple" Paseman, 2017.06.23. $\endgroup$ Jun 24, 2017 at 2:27
  • $\begingroup$ Thank you, I was hopping that there is a nice tool to do such problems. Do you have one in mind? $\endgroup$ Jun 24, 2017 at 3:07
  • $\begingroup$ This seems to prove that in fact all homotopy groups vanish, not just the first two. Is that right? $\endgroup$
    – roy smith
    Jun 24, 2017 at 15:59
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    $\begingroup$ Yes (which in the case at hand implies contractible). $\endgroup$ Jun 24, 2017 at 21:02

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