0
$\begingroup$

Let $M$ be a topological space.

Definition. $M$ is locally Euclidean of dimension $n$ if each point of $M$ has a neighborhood that is homeomorphic to an open subset of $\mathbb{R}^n$.

More specifically, if $M$ is locally Euclidean of dimension $n$, for each $p\in M$ we can find

$(i)\;$ an open subset $U\subseteq M$ containing $p$,

$(ii)\;$ an open subset $\hat{U}\subseteq\mathbb{R}^n$, and

$(iii)\;$ a homeomorphism $\varphi\colon U\to\hat{U}.$

Give the following characterization for the locally Euclidean spaces of dimension $n$:

Proposition. A topological space $M$ is locally Euclidean of dimension $n$ iff either of the following properties holds:

$(a)\;$ Every point of $M$ has a neighborhood homeomorphic to an open ball in $\mathbb{R}^n.$

$(b)\;$ Every point of $M$ has a neighborhood homeomorphic to $\mathbb{R}^n$.

Proof. $(\Longleftarrow)$ If $M$ has the property $(a)$ or $(b)$, then $M$ is locally Euclidean of dimension $n$, (open balls are open in $\mathbb{R}^n$ and $\mathbb{R}^n$ is open).

$(\Longrightarrow)$ Suppose that $M$ is locally Euclidean of dimension $n$. Since any open ball in $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$ itself, properties $(a)$ and $(b)$ are equivalent, so we need only prove $(a)$. Let $p\in M$, for hypotesis exists a neighborhood $U$ of $p$ and a homeomorphism $\varphi\colon U\to\hat{U}$, where $\hat{U}\subseteq\mathbb{R}^n$ is open. Since $\hat{U}$ is open, exists $r>0$ such that $B:=B(\varphi(p),r)\subseteq\hat{U}$. Since $\varphi$ is continuous, $B$ is open in $\hat{U}$(because $B=\hat{U}\cap B$, and $B$ is open in $\mathbb{R}^n$) and $p\in\varphi^{-1}(B)$, the inverse image $\varphi^{-1}(B)\subseteq U$ is a neighborhood of $p$ ($\varphi^{-1}(B)$ is open in $U$).Therefore $\varphi_{|\varphi^-{1}(B)\times B}\colon\varphi^{-1}(B)\to B$ is the homeomorphism from a neighborhood of $p\in M$ and an open ball in $\mathbb{R}^n.$ $\square$

Question 1. Is the previous proof correct and sufficiently detailed?

At this point we know that if $M$ is locally Euclidean of dimension $n$, for each $p\in M$ we can find

$(i)\;$ an open subset $U\subseteq M$ containing $p$,

$(ii)\;$ an open balls $B:=B(\varphi(p),r)\subseteq\mathbb{R}^n$, and

$(iii)\;$ a homeomorphism $\varphi\colon U\to B.$

Now i want that $\varphi(p)=0$ and $r=1$.

Consider the map $\psi\colon B\to B(0,r)$, definied as $p\mapsto p-\varphi(p)$. It is easy to convince oneself that this map is a homeomorphism. Moreover, consider the map $f\colon B(0,r)\to B(0,1)$ defined as $p\mapsto p/r$, it is a homeomorphism. Then $f\circ\psi\circ\varphi\colon U\to B(0,1)$ is an homeomorphism. We can now rewrite the definition of locally Euclidean of dimension $n$ as follows: $M$ is locally Euclidean of dimension $n$, if for each $p\in M$ we can find

$(i)\;$ an open subset $U\subseteq M$ containing $p$,

$(ii)\;$ a homeomorphism $\varphi\colon U\to B(0,1).$

Question 2. Is this procedure correct? Is there any mathematical error in this procedure? Any comments?

Thanks!

$\endgroup$
1
$\begingroup$

Regarding question (1), it's correct.

Regarding question (2), you've made a mistake by writing the translation map $\psi$ as $p\mapsto p-\varphi(p)$. It should be $k\mapsto k-\varphi(p)$. The other parts are fine.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.