The following is a proof that we can define a topological manifold $M$ in two different ways that are equivalent, one requiring the coordinate charts (or more specifically the homeomorphism) to carry open sets into arbitrary open sets, and the other requiring them to carry open sets onto open balls (and hence also $\mathbb{R}^n$ by transitivity of homeomorphism). One direction is obvious, so the below is a proof of the other direction.
Suppose $x$ is a point of a topological $n$-manifold $M$, and $U$ is an open nbhd (neighborhood) of $x$ such that there exists a homeomorphism $f$ from $U$ onto an open set $V \subset \mathbb{R}^n$. Then since $V$ is open, there exists an open ball $B \subset \mathbb{R}^n$ centered at $f(x)$ and contained in $V$. Now note $f^{-1}$ exists and is continuous, so we can restrict it to $B$ to get our desired homeomorphism.
Q1: I wanted to confirm that the last step relies on two facts, and it's not just immediate some other way:
- If $f: U \to V$ is a bijection, then so too is $f$ restricted to any subset of $U$.
- If $f: U \to V$ is a continuous map, then so too is $f$ restricted to any subset of $U$.
I guess these facts can be combined into:
- If $f: U \to V$ is a homeomorphism, then so too is $f$ restricted to any subset of $U$.
EDIT: This question is no longer:
The answers to all of your questions are Yes. You can directly check by definition.
For example, if $f:U\rightarrow V$ is bijective, then if you restrict $f$ to $W_1\subset U$, then $f:W_1\rightarrow f(W_1)$ is of course bijective. For $f|_{W_1}:W_1\rightarrow f(W_1):=W_2$,we have any open set in $W_2$ is of form $K\cap W_2$, where $K\subset V$ is open. Then we have $$ f^{-1}(K\cap W_2)=f^{-1}(K)\cap f^{-1}(W_2)=\text{(open set in $U$})\cap W_1 $$ By definition,$f|_{W_1}$ is continuous.
For your Q2:Yes. Essentially, it really comes down to the topological fact that interior structure and nbhd structure are equivalent.