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this is a topology question :

True or false and justify:

1) the subspace of a simply connected space is simply connected

2) the quotient of a simply connected space is simply connected


My thoughts:

1) False. Consider X = $R^2$ and Y = $S^1 \subseteq $ X. Then X is connected since $R^2$ is convex and any convex space is connected. We know that $\pi_1(S^1) = \mathbb{Z}$ and for a connected subspace $\mathbb{Z}$ we have $\pi_1( \mathbb{Z}$) = {0}. Then $S^1$ is not simply connected. So the statement is false.

2) Not sure how to do this. I tried the X = $R$ and Y = $Z$ and have X/Y $\cong$ $S^1$, which is path connected and therefore simply connected.

Could you check whether my part 1) is correct, and help me with 2)?

Thanks in advance!

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  • $\begingroup$ Quotient $[0,1]$ by identifying $\{0,1\}$. The quotient that you mentioned also shows it. You already said $S^1$ is not simply connected in your argument for part (1). $\endgroup$ – user551819 Apr 25 '18 at 15:37
  • $\begingroup$ @totoro Thank you! I didn't see that, guess my brain is not functioning well... $\endgroup$ – Liz Apr 25 '18 at 15:45
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Part 1 is OK. For an even simpler example, you can even consider $\{0,1\} \subset \mathbb{R}$. The line $\mathbb{R}$ is connected and simply connected, but the two-elements set $\{0,1\}$ is not even connected, let alone simply connected.

Your part 2 is good too. The group $\mathbb{R}$ is simply connected, and its quotient by the group $\mathbb{Z}$ is the circle (which is a particular case of a topological quotient), which is not simply connected.

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  • $\begingroup$ Thank you for the super fast response! I think your example given for part 1 is much simpler. $\endgroup$ – Liz Apr 25 '18 at 15:46

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