# Abstract Nonsense

## Locally Euclidean Spaces

We wish to begin discussing the concept of manifolds (the motivation and definition will be given in a subsequent post). We have  covered to a degree much more than is necessary all the concepts need to define a manifold except one. Namely, the concept of a space being “locally Euclidean”. Intuitively a locally Euclidean space of dimension $n$ is one that “looks like” $\mathbb{R}^n$ on a small enough scale. Consider our planet, it is locally Euclidean of dimension $n$ for although it is embedded in $\mathbb{R}^n$ to us it looks like $\mathbb{R}^2$ thus giving the age-old misconception that the world is flat. It is clear (intuitively) then that the unit sphere $\mathbb{S}^2$ is locally euclidean of dimension $n$ although it is not homeomorphic to any Euclidean space (since it is compact and any Euclidean space is not). Another concrete example is $(-\infty,0)\cup(0,\infty)$ with the usual topology. Clearly, given any point in this space we can find a neighborhood contained in either of these components which then clearly looks like $\mathbb{R}$. Once again this space is not homeomorphic to $\mathbb{R}$ since it is not connected.

Enough with the motivation, let’s move onto the formal definitions.

Locally Euclidean of Dimension $n$: Let $X$ be a topological space such that for every $x\in X$ there exists some neighborhood $U$ of $X$ which may embedded as an open set in $\mathbb{R}^n$. In, other words there exists some homeomorphism $\varphi:U\to V$ where $V\subseteq\mathbb{R}^n$ is open.

Chart: If $U\subseteq X$ is open and $\varphi:U\to\varphi(U)$ is a homeomorphism where $\varphi(U)\subseteq\mathbb{R}^n$ is open we call the ordered pair $(U,\varphi)$ a chart on $X$ and if $x\in U$ we call it a chart at $x$.

Remark: It is clear from the above that a space is locally Euclidean if and only if there is chart at every point of $X$.

We first show that $V$ being an open set may be replaced with either an open ball or $\mathbb{R}^n$ itself.

Theorem: Let $X$ be locally Euclidean, then given any $x\in X$ there exists some neighborhood $U$ of $x$ such that $U\approx B_\varepsilon(\varphi(x))$ for some $\varepsilon>0$.

Proof: Let $x\in X$ be arbitrary. Since $X$ is locally Euclidean there exists neighborhood $V$ of $x$ such that there exists some homeomorphism $\varphi:V\to O$ where $O\subseteq\mathbb{R}^n$ is open. But, since $O$ is open and $\varphi(x)\in O$ there exists some $B_\varepsilon(\varphi(x))\subseteq O$. It follows that if $U=\varphi^{-1}(B_{\varepsilon}(\varphi(x)))$ that $\varphi\mid U:U\to B_{\varepsilon}(\varphi(x))$ is a homeomorphism. The conclusion follows. $\blacksquare$.

Corollary: Since every open ball in $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$ the above is still valid if $B_{\varepsilon}(\varphi(x))$ is replaced with $\mathbb{R}^n$.

It turns out that the property of being locally Euclidean is inherited by open subspaces as is shown in the following:

Theorem: Let $X$ be a locally Euclidean space of dimension $n$ and $\mathcal{S}$ an open subspace of $X$, then $\mathcal{S}$ is locally Euclidean of dimension $n$.

Proof: Let $x\in\mathcal{S}$ be arbitrary. Then, by $X$‘s assumed local Euclidean property there exists some neighborhood $U$ of $x$ such that there exists some homeomorphism $\varphi:U\to O$. Clearly then $U\cap\mathcal{S}$ is an open neighborhood of $x$ in $\mathcal{S}$ and $\varphi\mid U\cap\mathcal{S}:U\cap\mathcal{S}\to\varphi\left(U\cap\mathcal{S}\right)$ is the desired homeomorphism. $\blacksquare$.

It is of course natural to ask under what conditions does a map preserve the property of being locally Euclidean. It should come as a surprise to no one that being locally Euclidean is invariant under homeomorphism. But, as with most topological properties there is a weaker condition for this to be true. But before we may give it in it’s fullest generality we need a new definition.

Local Homeomorphism: If $X,Y$ are topological spaces and $\varphi:X\to Y$ a continuous map such that for every $x\in X$ there exists some neighborhood $U$ of $x$ such that $\varphi\mid U:U\to\varphi(U)$ is a homeomorphism and $\varphi(U)$ is open in $Y$. If so, we say that $X$ and $Y$ are locally homeomorphic (symbolized by $X\overset{\text{loc.}}{\approx}Y$ and $\varphi$ is called a local homeomorphism.

Theorem: Let $\varphi:X\to Y$ be a local homeomorphism, then $\varphi$ is open.

Proof: Let $O\subseteq X$ be open and let $\varphi(x)\in\varphi(O)$. By assumption there exists some neighborhood $U$ of $x$ such that $\varphi(U)$ is open in $Y$ and $\varphi\mid U:U\to\varphi(U)$ is a homeomorphism. Clearly then we have that $O\cap U$ is an open subspace of $U$ and thus by assumption $\varphi\mid_U\left(O\cap U\right)$ is an open subset of $\varphi(U)$. But, since $\varphi(U)$ is a subspace we know that $\varphi(O\cap U)=\varphi(U)\cap V$ for some open set $V$ in $Y$. It follows that $\varphi\mid_U(O\cap U)$ is open in $Y$. But, this means that $\varphi\mid_U(O\cap U)$ is a neighborhood of $x$ contained in $\varphi(U)$. The conclusion follows. $\blacksquare$

Theorem: Let $X$ be a locally Euclidean space of dimension $n$. Then, if $\varphi:X\to Y$ is a surjective local homeomorphism, then $Y$ is a locally Euclidean space of dimension $n$.

Proof: Let $\varphi(x)\in Y$ be arbitrary. By assumption there exists some $U$ of $x$ such that $\varphi\mid_U :U\to\varphi(U)$ is a homeomorphism and $\varphi(U)$ is open in $Y$. But, by assumption there also exists some neighborhood $V$ of $x$, some open set $O$ in $\mathbb{R}^n$, and some $\psi:V\to O$ which is a homeomorphism. Clearly then $V\cap U$ is an open subspace of $U$ and thus using previous methods we see that $\varphi\mid_U(V\cap U)$ is a neighborhood of $x$ in $Y$. But, clearly since $U\cap V$ is open we also have that $\psi\mid_{U\cap V}:U\cap V\to\psi\left(U\cap V\right)$ is a homeomorphism and $\psi(U\cap V)\subseteq\mathbb{R}^n$ is open (this is clear it is an open subset of the open subspace $O$). It follows that $\varphi\mid U(V\cap U)$ is a neighborhood of $\varphi(x)$ and $\psi\mid_{U\cap V}\circ\left(\varphi\mid_{U\cap V}\right)^{-1}:\varphi(U\cap V)\to\psi(U\cap V)$ is a homeomorphism with an open subset of $\mathbb{R}^n$. The conclusion follows. $\blacksquare$.

Remark: A similar idea will appear later when we discuss overlap maps. Also, the idea of local homeomorphisms is much richer than what we have shown here. Maybe one day we will explore the concept in its entirety.

The next theorem should come as a surprise to no one.

Theorem: Let $X_1,\cdots,X_m$ be a finite number of topological spaces such that $X_k$ is locally Euclidean of dimension $n_k$ for each $k=1,\cdots,m$. Then, $\displaystyle \prod_{j=1}^{m}X_j$ under the product topology  is locally Euclidean of dimension $\displaystyle \sum_{j=1}^{n}n_j$.

Proof: Let $\bold{x}\in X$ be arbitrary. By assumption there exists a neighborhood $U_k$ of $\pi_k(\bold{x})$ and a homeomorphism $\varphi_k:U_k\to O_k$ where $O_k\subseteq \mathbb{R}^n_k$ is open. Clearly then $U_1\times\cdots\times U_m$ is a neighborhood of $\bold{x}$ and $O_1\times\cdots\times O_m$ and open subset of $\mathbb{R}^{n_1+\cdots+n_m}$. It follows from an old theorem that $\varphi_1\times\cdots\times\varphi_m:U_1\times\cdots\times U_m\to O_1\times\cdots\times O_m$ given by

$(x_1,\cdots,x_m)\to \{\varphi_1(x_1)\}\times\cdots\times\{\varphi_m(x_m)\}$

or equivalently (if it’s hard to picture)

$\left(\pi_1(\varphi_1(x_1))),\cdots,\pi_{n_1}(\varphi_1(x_1))),\cdots,\pi_1(\varphi_m(x_m))),\cdots,\pi_{n_m}(\varphi_m(x_m)))\right)$

is a homeomorphism (since it’s the product of homeomorphisms). The conclusion follows $\blacksquare$

We now discuss particularly nice theorem regarding a base that can be imposed on $X$. But, first a definition.

Euclidean Ball (E.B.): Let $X$ be a locally Euclidean space of dimension $n$ and let $U\subseteq X$ be open and such that $U\approx B$ where $B\subseteq\mathbb{R}^n$ is an open ball. Then, we call $U$ a Euclidean ball (E.B).

Our next theorem in effect says that the Euclidean balls are sufficient to describe $X$‘s topology.

Theorem: Let $X$ be a locally Euclidean space of dimension $n$ then $X$ has an open base consisting entirely of E.B.s

Proof: Let $x\in X$, we have by assumption there exists some neighborhood $U_x$ of $x$ such that $U_x\approx B_x$ for some open ball $B_x\subseteq\mathbb{R}^n$, and let $\varphi_x$ be the associated homeomorphism. Define then $\Omega_x=\left\{B_{\varepsilon}(\varphi_x(x)):B_{\varepsilon}(\varphi(x))\subseteq B_x\right\}$ and $\Lambda_x=\left\{\varphi_x^{-1}(\omega):\omega\in\Omega_x\right\}$. It is clear that each element of $\Lambda_x$ is an open subspace of $U_x$ but since $U_x$ is open we have that it is an open subspace of $X$. Furthermore, it is not to hard to see that given some $\varphi_x^{-1}(B_{\varepsilon}(\varphi_x(x)))$ that $\varphi\mid_{\varphi_x^{-1}(B_{\varepsilon}(x)))}:\varphi^{-1}(B_{\varepsilon}(\varphi(x)))\to B_{\varepsilon}(\varphi(x))$ is a homeomorphism. Thus, each element of $\Lambda_x$ is an E.B. Finally, let $\displaystyle \Lambda=\bigcup_{x\in X}\Lambda_x$. It is clear that $\Lambda$ is a collection of E.B.s in $X$ it remains to show that is a base.

To see this let $y\in X$ be arbitrary and $N$ any neighborhood of it. Letting the notation be as in the previous paragraph we have that $E_y\cap N$ is an open subspace of $E_y$ and so $\varphi_y(E_y\cap N)$ is an open subspace of $B_y$ and so there exists some $B_{\varepsilon}(\varphi_y(y))$ such that $B_{\varepsilon}(\varphi_y(y))\subseteq B_y$. It follows that $\varphi_y^{-1}(B_{\varepsilon}(\varphi_y(y)))\subseteq E_y\cap N\subseteq N$ and $\varphi_y^{-1}(B_{\varepsilon}(\varphi_y(y)))\in\Lambda$. The conclusion follows. $\blacksquare$

It is not surprising that most of the “local properties” of $\mathbb{R}^n$ are inherited by locally Euclidean spaces. While we will discuss most of them in subsequent posts we prove one here.

Theorem: Let $X$ be locally Euclidean of dimension $n$. Then, $X$ is first countable.

Proof: Let $x\in X$ be arbitrary and let $U_x,\varphi_x,B_x$ be as in the last theorem. Then, just as before there exists some $B_{\varepsilon}(\varphi_x(x))$ such that $B_{\varepsilon}(\varphi_x(x))$. By the Archimedean principle there exists some $N\in\mathbb{N}$ such that $\displaystyle \frac{1}{N}<\varepsilon$. Define

$\mathcal{N}=\left\{\varphi_x^{-1}\left(B_{\frac{1}{n}}(\varphi_x(x))\right):n\in\mathbb{N}-\{1,\cdots,N\}\right\}$

Clearly $\text{card }\mathcal{N}=\aleph_0$ so it remains to show that it is an open base at $x$. To do this let $N$ be any neighborhood of $x$. Clearly then $N\cap U_x$ is a neighborhood of $x$ contained in $U_x$. Thus, we have that $\varphi_x(U_x\cap N)$ is open in $B_x$ and so there exists some $\delta>0$ such that $B_{\delta}(\varphi_x(x))\subseteq \varphi(E_x\cap N)$. Thus, appealing to the Archimedean principle again we may find some $M\in\mathbb{N}-\{1,\cdots,N\}$ such that $\displaystyle \frac{1}{M}<\delta$. Clearly then $\varphi_x^{-1}\left(B_{\frac{1}{M}}(\varphi_x(x))\right)\subseteq E_x\cap N\subseteq N$ and since it’s also in $\mathcal{N}$ the conclusion follows. $\blacksquare$

This ends our discussion for now. Next time we will talk about manifolds.