Let $X$ be the set of all point $(x,y)\in\mathbb{R}^2$ such that $y=\pm 1$ and let $M$ be the quotient of $X$ by the equivalance relation generated by $(x,-1)\sim (x,1)$, that is $(x_1,y_1)\sim(x_2,y_2)$ iff $x_1=x_2$ and $x_1\ne 0$ or $x_1=x_2$ and $y_1=y_1$.
I must prove that $M$ is locally Euclidean.
If I consider the following sets
$$U_1=\{[(x,1)]\;|\;x\ne 0\}\cup\{[0,1]\}$$
and
$$U_2=\{[(x,-1)]\;|\;x\ne 0\}\cup\{[0,-1]\}$$
I can easily find a homeomorphism between $U$ and $\mathbb{R}$ and therefore show that $M$ is locally Euclidean.
I think my question is silly, please forgive me.
Question. $U_1$ and $U_2$ are open in $M$?
Thanks!
Let $\pi: X \to M$ the projection to the quotient, then $U_i$ is open in M iff $\pi^{-1}(U_i)$ is open in X.
We have to $\pi^{-1}(U_1)=X-\{(0,-1)\}$ and $\pi^{-1}(U_2)=X-\{(0,1)\}$ then both are open in X so $U_1,U_2$ are open in M.