Cfrgtkky

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 $varphi: U to V$

$ $

2. 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.

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




3. Why is $varphi(p)=0$?




4. 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)

5. 
   

   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 correct as Tu states it. Here's what happens with the main three doubts you have:

1) Why can we take $phi(p)=0$? By composing any homeomorphism $phi: Urightarrow V$ with the translation $vmapsto v-phi(p)$ (which is a homeomorphism of $mathbb{R}^n$) gives us a new $phi$ with $phi(p)=0$.

2) Why can we take $phi: Vrightarrow U=B(0,epsilon)$? The definition is that exists an open neighborhood $U$ of $p$, an open subset $Wsubset mathbb{R}^n$ containing $0$ by the step above, and a homeomorphism $phi: Urightarrow V$. The thing is that $W$ as open containing $0$, must contain a little ball $V$ centered at $0$. Now just rename $U=phi^{-1}(V)$ and you got the homeomorphism $phi: Urightarrow V=B(0,epsilon)$.

3) Why, in the case of the cross $C$, can we rule out the possibility that a neighborhood of $p$ in $C$ is homeomorphic to some open in $mathbb{R}^n$ with $ngeq 2$?
The thing is that in any topological manifold $M$, this $n$ is unique on each connected component of $M$. Otherwise, i.e if you had two $phi:Urightarrow Vsubset mathbb{R}^n$, $psi:Urightarrow Wsubset mathbb{R}^m$ homeomorphisms with $mneq n$, then you'd have that $psicircphi^{-1}: Vrightarrow W$ is a homeomorphism between open subsets of two different Euclidean spaces. This cannot be, due to a theorem called Invariance of Domain. For this part just check Spivak's A Comprehensive Introduction to Differential Geometry, Vol 1, Ch. 1.

Now with this in mind, it's clear that if our cross $C$ is locally Euclidean at $p$ (the intersection point), it's $n$ will have to be one. To see this, if the two lines crossing gives you trouble to conceal the $n$, just move a little away from $p$ on any of the lines, clearly this part is locally $mathbb{R}$. So the $n$ at p must also be one since the cross has just one component.

Now Tu's argument is clearer.

The definition of locally Euclidean is fine.

Suppose a point $pin M$ has an open neighbourhood $U$ such that there exists an open set $VsubseteqBbb R^n$ and a homeomorphism $varphi:Uto V$. 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 $vinBbb R^n$, the map $Bbb R^ntoBbb R^n, xmapsto x+v$ 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 $Vto V-varphi(p), xmapsto x-varphi(p)$, is a homeomorphism. The map $f:Uto V-varphi(p), f(x):=varphi(x)-varphi(p)$, 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.