Cross is not locally Euclidean in Tu Manifolds Tu Manifolds Section 5.1
Definition of locally Euclidean of dimension n.

Example


Firstly, to show the cross is not locally Euclidean


*

*Is this the definition that a space $M$ is locally Euclidean of some dimension?


$\exists n \in \mathbb N: \forall p \in M,$


*

*$\exists$ neighborhood $U$ of $p$ and in $M$,

*$\exists$ $V$ open in $\mathbb R^n$

*$\exists$ a homeomorphism $\varphi: U \to V$
$ \ $


*Can we show the cross $M$ is not locally Euclidean of any dimension, by showing the following?


$\forall n \in \mathbb N, \ \exists p \in M$ such that for all
$\forall n \in \mathbb N: \exists p \in M$:


*

*$\forall$ neighborhoods $U$ of $p$ and in $M$,

*$\forall$ $V$ open in $\mathbb R^n$

*$\forall$ maps $\varphi: U \to V$
$\varphi$ is not a homeomorphism.

Next, what is the proof exactly? Here is my attempt.

Let $n \in \mathbb N$. Choose $p$ to be the intersection. Let $U$ be any neighborhood of $p$ in $M$.
For some reason, $V$ is an open ball namely $V=B(0,\varepsilon)$.
Let $\varphi: U \to V$ be a map. For some reason $\varphi(p)=0$. If $\varphi$ were a homeomorphism, then $\varphi_R: U \setminus p \to B(0,\varepsilon) \setminus 0$ is a homeomorphism too, but this is a contradiction because of the facts about components.



*

*Tu does not give equivalent definitions of locally Euclidean so far, so is this one ball is enough?


I checked the appendices, and I did not find any such convention that "open set in $\mathbb R^n$" means element of basis of open balls.


*Why is $V=B(0,\varepsilon)$?

*Why is $\varphi(p)=0$?

*Here is what I did instead. Is this correct?

Let $n \in \mathbb N$. Choose $p$ to be the intersection. Let $U$ be any neighborhood of $p$ in $M$.
Case 1: $V$ is an open ball, $V=B(x,\varepsilon)$ for some $x \in \mathbb R^n$, not necessarily the origin.
Let $\varphi: U \to V$ be a map. If $\varphi$ were a homeomorphism, then $\varphi_R: U \setminus p \to B \setminus \varphi(p)$, where $\varphi(p)$ is not necessarily $0$ or $x$, is a homeomorphism too, but this is a contradiction because of the facts about components.
Case 2: $V$ is an open set but not an open ball.
I don't know this! (See the next question)



*How do we show the cross is not locally Euclidean for any open set $V$ in $\mathbb R^n$?


*

*$V$ is the union of basis elements, each of which are open balls and each of which are homeomorphic to a single open ball (because of what would be Part Two of this). I am now thinking about arcs or edges, so I know I am overthinking:

*I think we can show the cross is not locally Euclidean for any $V$ if any space $Y$ which is the union of open subspaces $\{A_{\alpha}\}$ and each $A_{\alpha}$ of which is homeomorphic to a single space $X$, is itself homeomorphic to $X$, and I think that would be true by pasting lemma, but I don't know how to address the part where $f=g$ on the intersections.
Pasting lemma from Munkres:


 A: The definition of locally Euclidean is fine.
Suppose a point $p\in M$ has an open neighbourhood $U$ such that there exists an open set $V\subseteq\Bbb R^n$ and a homeomorphism $\varphi:U\to V$. You may not necessarily have that


*

*$V$ is an open ball centred at origin,i.e. $V=B(0,\varepsilon)$ 

*$\varphi(p)$ is the origin, i.e. $\varphi(p)=0$,


but you can construct a new function from $\varphi$ satisfying the above two properties.
First, given a fixed vector $v\in\Bbb R^n$, the map $\Bbb R^n\to\Bbb R^n, x\mapsto x+v$ is a homeomorphism (and, it still is a homeomorphism upon restricting the domain and codomain properly). In particular, choose $v=-\varphi(p)$ and the map $V\to V-\varphi(p), x\mapsto x-\varphi(p)$, is a homeomorphism. The map $f:U\to V-\varphi(p), f(x):=\varphi(x)-\varphi(p)$, is a homeomorphism from an open neighbourhood of $p$ to an open set of $\Bbb R^n$, with the property that $p$ is sent to the origin $0$.
Second, restrict the domain and codomain of $f$ properly. The codomain is restricted to an open ball $B(0,\varepsilon)$ while the domain is restricted to its preimage $f^{-1}(B(0,\varepsilon))$. The map $f:f^{-1}(B(0,\varepsilon))\to B(0,\varepsilon)$ is a homeomorphism between the two sets. What you may want to prove is that $f^{-1}(B(0,\varepsilon))$ is an open neighbourhood of $p$.
So, my comment to your attempt, is that you don't need to prove case 2 at all, if you have proved for case 1 already.
