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 January 29th, 2019, 04:45 AM #1 Newbie   Joined: Jan 2019 From: italy Posts: 6 Thanks: 0 Fundamental group Hi 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? Last edited by skipjack; January 29th, 2019 at 06:10 AM.
 January 29th, 2019, 07:10 AM #2 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 312 Thanks: 111 Math Focus: Number Theory, Algebraic Geometry Hint: you might like $Y_1 \cup Y_2$ to be a circle. It's hard to say much more without giving the whole game away. Thanks from veronicalachi
 January 29th, 2019, 11:20 AM #3 Newbie   Joined: Jan 2019 From: italy Posts: 6 Thanks: 0 Maybe I have an idea. $S'−{p_0}$ (where $S'$ is the circonference with radius 1 and $p_0$ is the point with coordinates (1,0) ) is homeomorphic (through the projection function) to $\mathbb R$ so it has trivial fundamental group. The singleton ${p_0}$ has trivial fundamental group but S' has fundamental group isomorphic to Z. So I can take $Y_1=S'−{p_0}$ and $Y_2={p_0}$. Am I right?
January 29th, 2019, 12:22 PM   #4
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 Originally Posted by veronicalachi Maybe I have an idea. $S'−{p_0}$ (where $S'$ is the circonference with radius 1 and $p_0$ is the point with coordinates (1,0) ) is homeomorphic (through the projection function) to $\mathbb R$ so it has trivial fundamental group. The singleton ${p_0}$ has trivial fundamental group but S' has fundamental group isomorphic to Z. So I can take $Y_1=S'−{p_0}$ and $Y_2={p_0}$. Am I right?
Yes, this works, well done! One nitpick, though: $S'−{p_0}$ is indeed homeomorphic to $\mathbb{R}$, but not via projection (the projections of $S'−{p_0}$ to $\mathbb{R}$ are neither injective nor surjective).

 January 30th, 2019, 09:28 AM #5 Newbie   Joined: Jan 2019 From: italy Posts: 6 Thanks: 0 I meant the stereographic projection. Am I right? Thank you very much
January 30th, 2019, 09:37 AM   #6
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Quote:
 Originally Posted by veronicalachi I meant the stereographic projection. Am I right? Thank you very much
Ah of course, my mistake. That's all fine then!

 January 30th, 2019, 10:01 AM #7 Newbie   Joined: Jan 2019 From: italy Posts: 6 Thanks: 0 Thanks

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