I am trying to solve the following Problem in the book Introduction to smooth manifolds by Lee:
Let $X$ be the set of all points $(x,y) \in \mathbb{R}^2$ such that $y=\pm 1$, and let $M$ be the quotien of $X$ by the equivalence relation generated by $(x,-1) \sim (x,1)$ for all $x \neq 0$. Show that $M$ is locally Euclidean and second countavle, but not Hausdorff.
First I just wrote down that the quotientopology was: Quotientopology: Let $X$ be a topological space, and $\sim$ an equivalence relation and define $Y:=X/ \sim$. The map $q: X \rightarrow Y$ maps elements $x \in X$ to their equivalence classes $[x] \in Y$. Thus the set $S \subset Y$ is open if $q^{-1}(S)$ is open.
Locally Euclidean: We have to show that every point $m \in M$ has an open neighborhood that is homeomorphic to an open set in $\mathbb{R}$.
Well, after drawing a picture of what $M$ should look like, the identity map $id: Y \rightarrow X$ should be a homeomorphism.
So let $m \in Y$ and $m$ not be the origin. Then a neighborhood of $m $ looks like $[(-\epsilon+m,\epsilon+m)]$ so the identity $id:X \rightarrow Y$ is continuous, bijective and its inverse is continuous.
So let $m \in Y$ then a neighborhood of $m$ at the "origin" should look like $[(-\epsilon,\epsilon)]$, so $id$ should also be a homeomorphism.
second countable: Since $\mathbb{R}$ with it's topology induced by $|\cdot|$ is second countable. So the topology on $M$ is too. One can see that by taking finite open cover of $X$,call it $\mathcal{C}$. Then $\cup_{C \in \mathcal{C}}q(C)$ is an open covering of $M$.
The last step would be to show that $M$ is not Hausdorff. But I do not really understand why it is not Hausdorff.
Questions:
Why is $M$ not Hausdorff (I assume we would get a problem in the "origin" but I can't write it down)?
Looking at my argumentation for "locally Euclidean", something seems a little bit off. Is my proof of the locally Euclidean part correct?
