# 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

1. 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 $$\varphi: U \to V$$

$$\$$

1. 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.

1. 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.

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

2. Why is $$\varphi(p)=0$$?

3. 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)

1. 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:  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

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