# Exercise 2 in Hatcher, section 1.2: the union of convex sets is simply connected

Here is my problem :

I have open, convex subsets $$X_1, \dots, X_n$$ of $$\mathbb R^m$$ such that $$X_i \cap X_j \cap X_k$$ is never empty. I have to show that $$\bigcup\limits_{i=1}^nX_i$$ is simply connected.

This is the chapter about Van Kampen's theorem so I thought we have to use Van Kampen's theorem. Almost all the hypothesis are true.

But I just read that the intersection $$\bigcap\limits_{i=1}^nX_i$$ is not always non-empty, and a inside this intersection is needed for applying Van Kampen's. Did I make some mistakes? Maybe there is a smarter way to apply Van Kampen's Theorem here?

Thanks.

Do it by induction on $n$. The case $n = 2$ is exactly Van Kampen's theorem: $X_1$ and $X_2$ are path-connected, and so is their intersection (it's the intersection of two convex sets, hence convex, and it's nonempty, hence path-connected). Thus $$\pi_1(X_1 \cup X_2) \cong \pi_1(X_1) *_{\pi_1(X_1 \cap X_2)} \pi_1(X_2) = 0.$$

Now for the induction step, suppose that you know the result is true for a given $n \ge 2$ and let $X_1, \dots, X_{n+1}$ be convex sets such that $X_i \cap X_j \cap X_k$ is connected for all $i,j,k$. By the induction hypothesis, $Y = X_1 \cup \dots \cup X_n$ is simply connected. It remains to show that $Y \cap X_{n+1}$ is path-connected, and you can apply Van Kampen's theorem again to conclude. (1)

So let's show $Y \cap X_{n+1}$ is path-connected. Clearly $$Y \cap X_{n+1} = (X_1 \cap X_{n+1}) \cup \dots \cup (X_n \cap X_{n+1}).$$

Now suppose $x,y$ belong to $Y$, we are looking for a path from $x$ to $y$. Let $$x \in X_i \cap X_{n+1}, \qquad y \in X_j \cap X_{n+1}.$$ By the hypothesis on the $X_\cdot$, the intersection $X_i \cap X_j \cap X_{n+1}$ is non-empty; choose $z$ inside it. Since $X_i \cap X_{n+1}$ is path-connected (it's the nonempty intersection of two convex sets), there is a path $\gamma$ from $x$ to $z$. Similarly, there is a path $\gamma'$ from $z$ to $y$. Concatenating these two path gives a path $\alpha = \gamma \cdot \gamma'$ from $x$ to $y$. Thus $Y \cap X_{n+1}$ is path-connected and we can conclude (cf. (1)).

Remark. You forgot the assumption (which is written in Hatcher's book) that the sets $X_i$ have to be open. This is crucial to apply van Kampen's theorem, in general it's false.

• Perfect ! I was thinking to use directly Van Kampen's on $X_1, \dots, X_n$ but I was not able to prove that the intersection is not empty. Maybe they are counter-example. Thanks !
– user171326
Commented Mar 15, 2015 at 11:13
• @Najib I thought it was quite straightforward to prove. Since $X_i \cap X_j \cap X_k \neq \emptyset$ for all $i,j,k$, and each $X_{\alpha}$ is path connected, it follows that $X$ and intersections are path connected. It remains to be shown that $\pi_1(X) =0$. Van-Kampen's theorem yields that $$\pi_1(X) \cong \frac{\pi_1(X_1) \star \pi_1(X_2) \star \cdots \star \pi_1(X_n)}{N}.$$ We have that $\pi_1(X_{\alpha}) = 0$, however, given that for each $\alpha \in [1,n]$, $X$ is a convex set and therefore deformation retracts to a single point. It follows that $\pi_1(X)=0$ and $X$ is simply connected.
– user319128
Commented Aug 3, 2016 at 1:13
• There is a typo I think, after clearly it's $Y\cap X_{n+1}$. Van Kampen's theorem gives us that the fundamental group is trivial, it remain to prove that $X$ is path connected, isn't it ? Thanks (which is easy, just to be sure about the result)
– Alex
Commented May 16, 2017 at 10:37
• And we don't suppose that each $X_i$ are open sets, so we have find an open set $U_i$ such that $X_i\subset U_i$ and and homotopics. Right?
– Alex
Commented May 16, 2017 at 10:49
• @Alex Yes, there was a typo, thank you. It is proven (by induction) in my answer that $X$ is path-connected, reread it. And in Hatcher's book it was indeed specified that the $X_i$ are open, OP just forgot to write it down... Commented May 22, 2017 at 7:50

I think the fact that the cover is finite is not really needed here. Let's change $X_1,...,X_n$ to $\{X_i\}$. Let $i_0$ be a fixed index and $x_0\in X_{i_0}$. Define $Y_i=X_{i_0}\cup X_i$. Note that $Y_i$ is open path connected because $X_{i_0}$ and $X_i$ are path connected open sets with nonempty intersection. $Y_i\cap Y_j=X_{i_0}\cup(X_i\cap X_j)$ which is path connected because $X_{i_0}\cap X_i\cap X_j\neq \emptyset$ and both $X_{i_0}$ and $X_i\cap X_j$ are path connected. Each $Y_i$ is also simply connected by applying van Kampen to $\{X_i,X_{i_0}\}$. After checking all of these, we can apply van Kampen to $\{Y_i\}$ and thus conclude that $X$ is simply connected.