Let \(X = T^2 = S^1 \times S^1\) and \(X = U_1 \cup \cdots \cup U_n\) where each \(U_i\) is open and simply-connected. Show that there is an \(i\in\{2,\ldots,n\}\) for which \[ V_i := U_i \cap (U_1 \cup \cdots \cup U_{i-1}) \] is either empty or not path-connected. (Hint: argue by contradiction.)
Solution
For \(i=1,\ldots,n\), set \[ W_i=U_1\cup\cdots\cup U_i. \]
1. We must have \(n\ge2\).
If \(n=1\), then \[ T^2=X=U_1. \] But \(U_1\) is simply connected by hypothesis, whereas \[ \pi_1(T^2)\cong\mathbb Z^2\ne0. \] This is impossible.
2. Suppose, toward a contradiction, that for every \(i=2,\ldots,n\), the set\[ V_i=U_i\cap W_{i-1} \]is nonempty and path-connected.Then every \(W_i\) is simply connected.
We argue by induction on \(i\).
For \(i=1\), \[ W_1=U_1 \] is simply connected by hypothesis.
Now let \(i\ge2\) and assume \(W_{i-1}\) is simply connected. Both \(W_{i-1}\) and \(U_i\) are open in \(X\), and both are path-connected because they are simply connected. By the contradiction hypothesis, \[ W_{i-1}\cap U_i=V_i \] is nonempty and path-connected. Hence \[ W_i=W_{i-1}\cup U_i \] is path-connected.
Choose a basepoint in \(V_i\). The Seifert–van Kampen theorem gives \[ \pi_1(W_i) \cong \pi_1(W_{i-1})*_{\pi_1(V_i)}\pi_1(U_i). \] Both outer groups are trivial, since \(W_{i-1}\) and \(U_i\) are simply connected. Therefore \[ \pi_1(W_i)=0. \] Together with path-connectedness, this says that \(W_i\) is simply connected.
The induction is complete.
3. The assumption in 2 is impossible.
By 2, \[ W_n \] is simply connected. But the \(U_i\) cover \(X\), so \[ W_n=X=T^2. \] Thus \(T^2\) would be simply connected, contradicting \[ \pi_1(T^2)\cong\mathbb Z\oplus\mathbb Z. \]
4. Consequently, for some \(i\in\{2,\ldots,n\}\), the set \(V_i\) is empty or is not path-connected.
This is the negation of the impossible assumption in 2.