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January 29th, 2019, 05:45 AM   #1
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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 07:10 AM.
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January 29th, 2019, 08:10 AM   #2
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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.
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January 29th, 2019, 12:20 PM   #3
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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?
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January 29th, 2019, 01:22 PM   #4
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Quote:
Originally Posted by veronicalachi View Post
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).
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January 30th, 2019, 10:28 AM   #5
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I meant the stereographic projection. Am I right? Thank you very much
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January 30th, 2019, 10:37 AM   #6
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I meant the stereographic projection. Am I right? Thank you very much
Ah of course, my mistake. That's all fine then!
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January 30th, 2019, 11:01 AM   #7
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