Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a topological space and $U,V \subset X$ two open subsets such that $U \cap V$ and $U \cup V$ are both simply connected. How can i show that $U$ and $V$ are simply connected? Thanks in advance.


share|cite|improve this question
yes i know. but i tried with van kampen's. but i have no idea how to apply it. – hilary Dec 9 '11 at 8:15

Ok. Let me expand my answer. By Seifert–van Kampen theorem, we know that the fundamental group $\pi_1(U\cup V)$ of $U\cup V$ is the free product of the fundamental groups of $U$ and $V$ with amalgamation of $\pi_1(U\cap V)$. Since $U\cap V$ is simply connected by assumption, i.e. $\pi_1(U\cap V)=0$, $\pi_1(U\cup V)$ is the free product of $\pi_1(U)$ and $\pi_1(V)$. Since $U\cup V$ is simply connected by assumption, i.e. $\pi_1(U\cup V)=0$, we must have $\pi_1(U)=0$ and $\pi_1(V)=0$; otherwise, if $\pi_1(U)\neq 0$ or $\pi_1(V)\neq 0$, the free product of $\pi_1(U)$ and $\pi_1(V)$ must be non-trivial. For the above facts about free product, you can refer to here.

Note added: As Chris said, I should prove that $U$ and $V$ are path-connected before I can apply Seifert-van Kampen theorem. Here is the proof: note that $U\cup V$ and $U\cap V$ are simply-connected by assumption, which implies that $U\cup V$ and $U\cap V$ are connected. Now by the Mayer-Vietoris sequence, we have $$H_0(U\cap V)\rightarrow H_0(U)\oplus H_0(V)\rightarrow H_0(U\cup V)\rightarrow 0.$$ Since the rank of the zero homology $H_0$ is equal to the number of connected components, by the above exact sequence $H_0(U)$ and $H_0(V)$ has rank 1, which implies that $U$ and $V$ are connected. Since $U$ and $V$ are open by assumption, $U$ and $V$ must be path-connected.

share|cite|improve this answer
Surely we need to show that $U$ and $V$ are path-connected before we can apply the theorem? – Chris Eagle Dec 9 '11 at 14:54
Right, but I think this is standard: if $U$ and $V$ are simply connected, then of course they are connected. Moreover, they are open by assumption. And it can be proved that open connected set must be path connected. – Paul Dec 10 '11 at 1:15
But we need $U$ and $V$ to be path-connected before we can use van Kampen to conclude they are simply-connected. It's not hard to show that if $U$ and $V$ are open and both $U \cup V$ and $U \cap V$ are path-connected, then $U$ and $V$ are path-connected, but it has to be done. – Chris Eagle Dec 10 '11 at 1:28
@Chris: Oh yes, you are right. I should be more careful. See my edited answer. – Paul Dec 10 '11 at 5:55
Please do not overlook the fact that there is a Seifert-van Kampen theorem for the fundamental groupoid on a set of base points, which allows for the computation of the fundamental group of the circle, as well as many other examples. For the use of groupoids, see – Ronnie Brown Apr 26 '12 at 21:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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