# Is the plane with two origins a manifold?

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)?

• The open neighbourhoods are like balls, no extra origin is in it. Commented Nov 20, 2021 at 13:55
• 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 $B$. Commented Nov 20, 2021 at 13:57

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 $$X$$).