Exercise 1.1 (Equivalent definitions of topological manifolds)

Proof. We use a circular chain of implications to demonstrate the equivalence of the three definitions.

( $1⟹2$)

Suppose that $M$ is a smooth manifold, i.e., each point $p$ of $M$ has a neighborhood $U$ that is homeomorphic to an open subset $Û$ of ${ℝ}^n$ via the homeomorphism $\varphi :U\to Û$. Since $Û$ is open, we can find an open ball $B(f(p),r)$, $r>0$ small enough, around $f(p)$ such that $B(f(p),r)\subseteq Û$. Then $U^{\prime }:={\varphi }^{-1}B(f(p),r)$ is an open subset of $U$ containing $p$. As such, $\varphi {\text{↾}}_{B(f(p),r)}$ is a homeomorphism from an open neighbourhood $U^{\prime }$ of $p$ to the open ball $B(f(p),r)$. In other words, every point $p\in M$ has a neighborhood that is homeomorphic to an open ball in ${ℝ}^n.$

( $2⟹3$)

Now suppose that each point in $M$ has neighbourhood $U$ that is homeomorphic to an open ball $B(x_0,r)$, $x\in {ℝ}^n$, $r>0$, in ${ℝ}^n$. Since all open balls are homeomorphic to the unit ball $B(0,1)$ via the homeomorphism of transition and scaling

$${\varphi }_{(x_0,r)}:B(x_0,r)\to B(0,1),x\mapsto \frac{x-x_0}{r},$$

$U$ must be homeomorphic to $B(0,1)$. But the unit ball $B(0,1)$ itself is homeomorphic to the Euclidean space via the bijection

$$\varphi :B(0,1)\to {ℝ}^n,x\mapsto \frac{x}{1-\parallel x\parallel }.$$

Thus, by transitivity, the neighborhoods of points in $M$ are homeomorphic to all of the ${ℝ}^n$.

( $3⟹1$)

Finally, suppose that every point $p$ in $M$ has a neighborhood $U$ which is homeomorphic via $\varphi :U\to {ℝ}^n$ to the whole of ${ℝ}^n$. Any open subset $Û\subseteq {ℝ}^n$ results in the open preimage $U^{\prime }:={\varphi }^{-1}(Û)$ under $\varphi$; thus, the restriction ${\varphi }^{\prime }:=\varphi {\text{↾}}_{U^{\prime }}$ of $\varphi$ under $U^{\prime }$ is a homeomorphism from the neighborhood $U^{\prime }\subseteq U$ of $p$ to an open subset $Û$ of ${ℝ}^n$.