# On the locally Euclidean topological spaces of dimension n

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!

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.