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I was reading the definition of topological 2-manifold and since one asks for the space to be Hausdorff and locally homeomorphic to $\mathbb{R}^{2}$, I wanted to find some space $X$ satisfying the latter condition but not Hausdorff. Following the idea of the line with two origins, I wondered if I could get such $X$ doing the same in $\mathbb{R}^{2}$, namely, considering $\mathbb{R}^{2}\times\{0\}\sqcup\mathbb{R}^{2}\times\{1\}$ and taking the quotient by the relation $(x,y,0)\sim(x,y,1)$ iff $(x,y)\neq(0,0)$. Then I see that given a point that isn't one of the two origins there is an open neighborhood homeomorphic to the open ball in $\mathbb{R}^{2}$ (for the quotient just gives $\mathbb{R}^{2}$ except at the origins), but then if $p$, $q$ are the origins in the quotient I don't see how to get a neighborhood of $p$ homeomorphic to the open ball (since the open neighborhoods are like balls in $\mathbb{R}^{2}$ but with an extra origin and that doesn't feel homeomorphic to an open ball). Is there a way to prove there is such a neighborhood, or am I just wrong (and in that case, is there an example of the space $X$ I want to find)?

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  • $\begingroup$ The open neighbourhoods are like balls, no extra origin is in it. $\endgroup$ Commented Nov 20, 2021 at 13:55
  • $\begingroup$ Check that $B \times \{0\} \sqcup B\setminus \{(0,0)\} \times \{1\}$ is saturated for $\sim$ and open in the sum space so corresponds to a homeomorphic set to $B$. $\endgroup$ Commented Nov 20, 2021 at 13:57

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If we have quite generally for any $X$, the quotient $X_p:= (X \times \{0\} \sqcup X \times \{1\}){/}R_p$ where $R_p$ is the equivalence relation generated by $$(x,0) \sim (x,1) \text{ if } x \neq p$$ for some fixed $p \in X$, then if $X$ is $T_1$ the class $[(p,0)]$ has a basic neighbourhood $q[(U \times \{0\}) \sqcup (U \times \{1\}\setminus \{(p,1)\})$ which is just homeomorphic to $U \subseteq X$ (and similarly for $[(p,1)]$ (where $U$ is an open neighbourhood of $p$ in $X$). So e.g. the plane with two origins is locally Euclidean. But $[(p,0)]$ and $[(p,1)]$ do not have disjoint neighbourhoods so $X_p$ is not Hausdorff (if $p$ is not isolated in $X$).

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