find two topological spaces $Y_1$, $Y_2$ such that $\pi (Y_1)=\pi (Y_2)= (0)$ but $\pi (Y_1 \cup Y_2)\ne (0)$

Question

I everybody, I have to solve this exercise. I have to find two topological spaces $Y_1$, $Y_2$ such that $\pi (Y_1)=\pi (Y_2)= (0)$ but $\pi (Y_1 \cup Y_2)\ne (0)$. Can you help me?

Answer 1

Note that $\mathbb{S}^1 \setminus \{(1, 0)\}$ is homeomorphic to $\mathbb{R}$, so $\pi_1(\mathbb{S}^1 \setminus \{(1, 0)\}) \cong \pi_1(\mathbb{R}) = (0)$ and also $\pi_1(\{(1, 0\}) = (0)$, but $\pi_1(\mathbb{S}^1 \setminus \{(1, 0)\} \cup \{(1, 0)\}) = \pi_1(\mathbb{S}^1) = \mathbb{Z}$.

Answer 2

Hint: Take a space you know has nontrivial fundamental group and cover it with two simply connected subsets.