Does the subspace / quotient of a simply connected space always be simply connected?

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）？

• 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). – user551819 Apr 25 '18 at 15:37
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.