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 January 29th, 2019, 05:45 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 07:10 AM. January 29th, 2019, 08:10 AM #2 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 313 Thanks: 112 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, 12:20 PM #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, 01:22 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, 10:28 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, 10:37 AM   #6
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 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, 11:01 AM #7 Newbie   Joined: Jan 2019 From: italy Posts: 6 Thanks: 0 Thanks  Tags fundamental, group, topology Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post fromage Abstract Algebra 0 December 4th, 2016 03:39 AM Hayato Topology 3 September 8th, 2015 04:52 PM msv Abstract Algebra 1 February 19th, 2015 12:19 PM TTB3 Real Analysis 2 April 19th, 2009 09:52 AM mingcai6172 Real Analysis 0 March 21st, 2009 03:35 PM

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