Let \(X \subset \mathbb{R}^m\) be the union of convex open sets \(X_1, \cdots, X_n\) such that \(X_i \cap X_j \cap X_k \neq \emptyset\) for all \(i, j, k\). Show that \(X\) is simply-connected.
Solution
For \(1\le r\le n\), set \[ U_r=X_1\cup\cdots\cup X_r. \]
1. Each \(X_i\) is path connected and simply connected.
Each \(X_i\) is convex. For any two points \(x,y\in X_i\), the straight segment \[ (1-t)x+ty, \qquad 0\le t\le1, \] lies in \(X_i\), so \(X_i\) is path connected. The same straight-line homotopy contracts \(X_i\) to any chosen point of \(X_i\), so \(X_i\) is contractible and hence simply connected.
2. For every \(r\ge2\), the intersection\[ U_{r-1}\cap X_r = \bigcup_{i<r}(X_i\cap X_r) \]is path connected.
Each set \(X_i\cap X_r\) is convex, hence path connected. If \(i,j<r\), then \[ (X_i\cap X_r)\cap(X_j\cap X_r) = X_i\cap X_j\cap X_r, \] which is nonempty by hypothesis. Thus the family \[ \{X_i\cap X_r:i<r\} \] consists of path-connected sets with pairwise nonempty intersections. Their union is therefore path connected.
3. If \(U_{r-1}\) is simply connected, then \(U_r\) is simply connected.
The sets \(U_{r-1}\) and \(X_r\) are open in \(X\). By induction hypothesis, \[ \pi_1(U_{r-1})=0, \] and by 1, \[ \pi_1(X_r)=0. \] Their intersection is path connected by 2.
The Seifert–van Kampen theorem therefore gives \[ \pi_1(U_r) \cong \bigl(\pi_1(U_{r-1})*\pi_1(X_r)\bigr)/N =0, \] where \(N\) is the normal subgroup generated by the usual overlap relations. Since both free-product factors are trivial, the quotient is trivial.
The union \(U_r\) is path connected because it is the union of the path-connected sets \(U_{r-1}\) and \(X_r\) with nonempty intersection. Hence \(U_r\) is simply connected.
4. The space \(X=U_n\) is simply connected.
1.Repeated application of 3 gives simple connectivity of\[ U_2,U_3,\dots,U_n=X. \]