2
$\begingroup$

If $U$ is an open and connected subset in $\mathbb{R}^2$, then it is path connected. In further, we assume that $U$ is in a unit ball.

If its complement $U^c$ is connected, then show that $U$ is simply connected.

[Add] Definition : $X$ is simply connected if there is a continuous map $h: D^2\rightarrow X$ with $h|\partial D^2=c$ when $c: S^1\rightarrow X$ is continuous map.

$\endgroup$
  • 1
    $\begingroup$ It is necessary. Consider $U = \Bbb R^2 \setminus \{(0,0)\}$ $\endgroup$ – Kenny Lau Mar 29 '18 at 6:27
  • 1
    $\begingroup$ The condition (A) is necessary. Let $U$ be the set of points with distance to the origin larger than $1$. Then $U^c$ is the closed unit ball and connected, but $U$ is not simply connected. $\endgroup$ – Joppy Mar 29 '18 at 6:28
  • $\begingroup$ @Lau, Joppy : I see. Thank you. $\endgroup$ – HK Lee Mar 29 '18 at 6:28
  • 1
    $\begingroup$ With your definition of simply connected, you're basically forcing us to use homology... (or else how would you prove that $\Bbb R^2$ is simply connected, i.e. Jordan curve theorem?) $\endgroup$ – Kenny Lau Mar 29 '18 at 6:31
  • $\begingroup$ From $h(s,t)=sc(t)$, $\mathbb{R}^2$ is simply connected. $\endgroup$ – HK Lee Mar 29 '18 at 6:43
1
$\begingroup$

I'm not sure if this should be an answer or a comment, since it is an idea, or "sketch of proof ".

Consider the equivalent definition: $X$ is a simply connected space if it is path-connected and for all $x\in X$, any loop with basepoint $x$ (i.e. continuous $\alpha : I \rightarrow X$ such that $\alpha (0)=\alpha (1)=x$) is homotopy equivalent to the constant path $x$.

Suppose $U$ is not simply connected and take a simple closed curve (this is a loop), not homotopy equivalent to a constant path. This curve contained in $U$ divides the plane in two regions $A$ and $B$, one of them bounded, say $A$. The interior of $A$ has points of $U^c$, otherwise you could shrink the loop to a point staying within $U$. Observe that $B\cap U^c \neq \varnothing$ as well, since $U$ is bounded.

Now to show that $U^c$ is not connected, consider the pair of disjoint nonempty open sets of $U^c$ given by taking intersection of $U^c$ and the interiors of $A$ and $B$.

$\endgroup$
  • $\begingroup$ Since the comments in the question point to Jordan curve theorem and this answer is not explicit in what is being used, I must say that it uses the Schoenflies theorem implicitly, not only the JCT. $\endgroup$ – Aloizio Macedo Mar 31 '18 at 17:56
1
$\begingroup$

Consider $U$ inside the Riemann sphere. Since $U$ is contained in the unit ball, its complement $U^c$ in the complex plane remains connected when we add the point at infinity (i.e., when we consider the complement in the Riemann sphere instead). From now on, we then shift the problem to $U$ an open subset of a Riemann sphere which has as complement a connected set.

By Poincaré-Alexander duality, we then have that

$$H_1(U) \simeq \widetilde{\check{H}}{}^0(U^c) \simeq 0.$$

Since open subsets of the plane have free fundamental groups (free groups, not free abelian) and $H_1$ is the abelianization of the fundamental group, it follows that $U$ is simply connected (the abelianization of a free group with $\lambda$ generators is the free abelian group with $\lambda$ generators. Thus, if the abelianization is $0$, then so is the original).

OBS: $0$-th Cech-cohomology counts quasicomponents of the space. Since $U^c$ is compact and Hausdorff, this coincides with the components.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.