I'm having trouble solving one part of one of the initial exercises of the classic Boothby book "An Introduction to Differentiable Manifolds and Riemannian Geometry" (exercise I.3.1). To be more specific, it's asked to prove that $X = A \cup B\cup C$, where $A=\lbrace(x,y):x\geq 0, y=1\rbrace \subset \mathbb{R^{2}}$, $B=\lbrace (x,y):x\geq 0, y=-1\rbrace \subset\mathbb{R^{2}}$ and $C=\lbrace (x,y) : x<0,y=0\rbrace\subset\mathbb{R^{2}}$ is locally Euclidian but not a manifold. The topology considered is the subspace topology for $C$,$A\setminus(0,1)$ and $B\setminus(0,-1)$. For $(0,1),(0,-1)$ we have as basis of neighborhood the following sets respectively: $\lbrace (x,1):0\leq x <\delta\rbrace \cup \lbrace(x,0):-\delta\leq x <0\rbrace$ and $\lbrace (x,-1):0\leq x <\delta\rbrace \cup \lbrace(x,0):-\delta\leq x <0\rbrace$, with $\delta > 0$.
I proved that this cannot be Hausdorff and that it is locally Euclidian around any point of $X$ except $(0,1),(0-1)$. BUT, I cannot find a local homeomorphism between an open set of $\mathbb{R}$ and a neighborhood of $(0,1)$ or $(0,-1)$.
Can someone help me please ? Many thanks
Let's consider only the point $p = (0,1) \in A$, the case for $(0,-1) \in B$ is analogous.
If we denote the given basis neighbourhood $[-\delta,0[\times \{0\} \cup [0,\delta[\times \{1\}$ of $p$ by $B_\delta$, the first thing to note is that these neighbourhoods are not open in $X$. Indeed, the point $(-\delta,0)$ has a neighbourhood basis consisting of the sets $U_\eta = ]-\delta-\eta,-\delta+\eta[\times\{0\}$ for small enough positive $\eta$ since on $C$, the topology is the subspace topology, and none of the $U_\eta$ is contained in $B_\delta$.
Let's denote by $N_\varepsilon$, for $\varepsilon > 0$ the neighbourhood $]-\varepsilon,0[\times \{0\} \cup [0,\varepsilon[\times\{1\}$ of $p$. It is a neighbourhood of $p$ since $B_{\varepsilon/2} \subset N_\varepsilon$, and it is indeed an open neighbourhood of $p$, since $N_\varepsilon \cap C$ and $N_\varepsilon \cap (A\setminus\{p\})$ are both open in the subspace topology, so all points in $N_\varepsilon\setminus\{p\}$ are also interior points.
Then, for every $\varepsilon > 0$, the map
$$\varphi_\varepsilon \colon N_\varepsilon \to ]-\varepsilon,\varepsilon[; \quad \varphi_\varepsilon((x,y)) = x$$
is a homeomorphism.
To show that $\varphi_\varepsilon$ is continuous in any point $(x,y) \in N_\varepsilon$, we distinguish the three cases $x < 0$, $x > 0$ and $x = 0$.
For $x < 0$, we have $\varphi_\varepsilon^{-1}(]x-\delta,x+\delta[) = ]x-\delta,x+\delta[\times \{0\}$ for $0 < \delta < \min \{ \lvert x\rvert, \varepsilon - \lvert x\rvert\}$, and that is a neighbourhood of $(x,0)$ in $X$.
For $x > 0$ and $0 < \delta < \min \{x, \varepsilon - x\}$, we have $\varphi_\varepsilon^{-1}(]x-\delta, x+\delta[) = ]x-\delta,x+\delta[\times \{1\}$, and that is a neighbourhood of $(x,1)$ in $X$.
For $x = 0$ and $0 < \delta < \varepsilon$, we have $\varphi_\varepsilon^{-1}(]-\delta,\delta[) = N_\delta$, and that is also a neighbourhood of $p = (x,1)$ in $X$.
So $\varphi_\varepsilon$ is continuous. Its inverse $\psi_\varepsilon$ is also continuous, since