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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

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  • $\begingroup$ The basis neighbourhoods are homeomorphic to $(-\delta,\delta)$ via projection to the $x$-coordinate. $\endgroup$ – Daniel Fischer Oct 1 '13 at 17:28
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    $\begingroup$ You mean if i do the projection between $[-\delta,0[\times\lbrace 0\rbrace \cup [0,\delta[\times\lbrace 1\rbrace$ and $]-\delta,\delta[$ ? Why is the inverse of this application still continuous, considering the topology of X? $\endgroup$ – KLMN Oct 1 '13 at 17:42
  • $\begingroup$ Since it's a basis of neighborhood, I can have for some $\delta_{2}$ the following open set : $[-\delta_{2},0[\times\lbrace 0\rbrace \cup [0,\delta_{2}[\times\lbrace 1\rbrace \subset [-\delta,0[\times\lbrace 0\rbrace \cup [0,\delta[\times\lbrace 1\rbrace$, which the inverse of the projection is not an open set of $\mathbb{R}$. What am I seeing wrong ? $\endgroup$ – KLMN Oct 1 '13 at 17:47
  • $\begingroup$ And what is the image of the projection of the point $(\delta,0)$ in the interval $] -\delta,\delta[ $ ? $\endgroup$ – KLMN Oct 1 '13 at 17:51
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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

  • $\psi_\varepsilon^{-1}(]x-\delta,x+\delta[\times\{0\}) = ]x-\delta,x+\delta[$ is a neighbourhood of $x$ for $x < 0$ and $0 < \delta < \min \{ \lvert x\rvert, \varepsilon - \lvert x\rvert\}$,
  • and $\psi_\varepsilon^{-1}(]x-\delta,x+\delta[\times\{1\}) = ]x-\delta,x+\delta[$ is a neighbourhood of $x$ for $x > 0$ and $0 < \delta < \min \{ x, \varepsilon - x\}$,
  • and $\psi_\varepsilon^{-1}(N_\delta) = ]-\delta,\delta[$ is a neighbourhood of $0$ if $0 < \delta < \varepsilon$.
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  • $\begingroup$ Thank you Daniel, but in my book, the basis of neighborhoods is defined as $[\delta,0[\times\lbrace 0\rbrace \cup (...)$ and not as $]\delta,0[\times\lbrace 0\rbrace \cup (...)$. Doesnt it make a difference to prove that the inverse is continuous ? $\endgroup$ – KLMN Oct 1 '13 at 18:41
  • $\begingroup$ Ah, sorry. I thought that was a typo in your comments, haven't seen it in the question body. I assumed it was $]-\delta,0[$, because $[-\delta,0[$ doesn't make much sense. It doesn't change the facts, but I have to elaborate on that. Hang on, will take a few minutes. $\endgroup$ – Daniel Fischer Oct 1 '13 at 19:00
  • $\begingroup$ Updated. The main part, if I interpret your problems no correctly, is that the given neighbourhood basis of the unusual two points does not consist of open sets. If you choose an open neighbourhood basis, you get the more or less natural thing. $\endgroup$ – Daniel Fischer Oct 1 '13 at 19:15
  • $\begingroup$ AHHHH ! That "small" detail. I thought that by definition a basis of neighborhood consists of subsets that are also opens sets. Thank you very much for the wonderful and clear answer. $\endgroup$ – KLMN Oct 1 '13 at 19:24
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    $\begingroup$ No, openness is not required. Sometimes one wants for example a neighbourhood basis of compact neighbourhoods, and those are often closed but not open. Here, however, I suspect a plain typo. It could have been intentional, then it would be a mean trick to make things more difficult than they should be. But typo seems more likely. $\endgroup$ – Daniel Fischer Oct 1 '13 at 19:30

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