$\pi_1([S^2-\{(0,0,1),(0,0,-1)\}]/x\sim -x)$

Question

1. I'm interested in calculating $\pi_1(X)=\pi_1([S^2-\{(0,0,1),(0,0,-1)\}]/x\sim -x)$
2. Moreover, I'm interested in $Y=[S^2\cap \{x\in\mathbb R^3:\:x_1 x_2 x_3>0\}]/x\sim -x$.

Appreciate any kind help!

Answer 1

HINTS try to show that $X = \mathbb{RP^2}-x$ where $x$ is the image of (0,0,1) under the quotient map $q:S^2 \rightarrow \mathbb{RP^2}$ ...from here you can easily calculate the fundamental group of $X$ is $\mathbb{Z}$...(basically if you collapse attach a disc with the boundary of a mobius band then you will get $RP^2$ so intuitively if look at $RP^2$\a point it is difformation retract to mobius band which is difformation retract to $S^1$

in case of $Y$ observe that $x_3 >0$ so if $x \in Y$ then $-x \notin Y$...so basically $Y \cap X$ is a open hemisphere minus a point which is difformation retract onto a circle...so fundamental group is $\mathbb{Z}$...

now observe that if we attach a disc with the boundary of a mobius band we'll get $RP^2$... now $X$\ $Y$ is mobius band...now consider $X_1$ which is constructed by $X$ witha $\epsilon$ thinken boundary inside $Y$...so clearly $X_1$ d.r onto $X$ ( I just define $X_1$ so that rest of the argement makes sense)...if we consider $a \in \pi_1(X_1\cap Y)$ be the generator then $a$ is homotpic to the boundary of $X_1$ which is homotopic to the square of the generator of $X_1$...so $f_*:\pi_1(X_1\cap Y) \rightarrow \pi_1(X_1)$ as $z \mapsto z^2$..( sorry if you think the last argument is very intuitive)