Let's look at the circle $S^1=\{(x,y)\in\Bbb R^2\mid x^2+y^2=1\}$ and the map $f:\Bbb R\to\Bbb R^2, f(t):(\cos(t),\sin(t)).$
A subset $A\subseteq S^1$ is said to be good if $f^{-1}(A)$ is open in $\Bbb R$. Prove that good sets generate a topology and that that topology is equal to the relative topology on $S^1$ as a subset of $\Bbb R^2$ with the standard topology.
My attempt:
Let $G:=\{A\subseteq S^1\mid f^{-1}(A)\text{ is open in }\Bbb R\}.$
$S^1,\emptyset\in G$ because $f^{-1}(S^1)=\Bbb R$ and $f^{-1}(\emptyset)=\emptyset$ are open in $\Bbb R.$$\tag 1$
Let $I$ be some set of indices and $A_i\in G,i\in I$. Then, $f^{-1}(A_i)=V_i$ is open in $\Bbb R$. We have
$$\tag 2 f^{-1}\left(\bigcup_{i\in I} A_i\right)=\bigcup_{i\in I} f^{-1}(A_i)=\bigcup_{i\in I} V_i\quad\text{
is open in }\Bbb R\implies \bigcup_{i\in I}A_i\in G.$$
Let $k\in\Bbb N.$ We then have $$\tag 3 f^{-1}\left(\bigcap_{i=1}^k A_i\right)=\bigcap_{i=1}^kf^{-1}(A_i)=\bigcap_{i=1}^k V_i\quad\text{ is open in }\Bbb R\implies \bigcap_{i=1}^k A_i\in G.$$
Therefore, $G$ is a topology.
Now, let $U\subseteq\Bbb R^2$ be some open set.
$f(\Bbb R)=S^1$ is closed in $\Bbb R^2.$ Let $s\in S^1$.
$\forall r>0,$ we have $B(s,r)\nsubseteq S^1,$ so, I think $f^{-1}(U)=f^{-1}(A\cap U).$
Let $\tau$ denote the relative topology on $S^1$.
Since $f$ is continuous on $\Bbb R,$ for all open $U\subseteq\Bbb R^2, f^{-1}(U)$ is open in $\Bbb R$. Therefore, $f^{-1}(A\cap U)=f^{-1}(U)$ is open, hence $A\cap U\in\tau$ is also a good set, that is $A\cap U\in G.$
Since $U\subseteq\Bbb R^2$ was arbitrary and therefore, $A\cap U\in\tau$ as well, we have $\tau\subseteq G.$
We also need the other inclusion, $G\subseteq\tau,$ but, I don't know how to start with it. Are there any fallacies in my prior reasoning and how should I continue?
Best Answer
Your proof of $\tau \subset G$ is essentially correct, although your notation and reasoning is somewhat unclear.
What is $A$ in $f^{-1}(U)=f^{-1}(A\cap U)$? You certainly mean $f^{-1}(U)=f^{-1}(S^1\cap U)$ which is correct. In fact, for each function $\phi : X \to Y$ between sets $X,Y$ we have $f^{-1}(U) = f^{-1}(f(X) \cap U)$ for all $U \subset Y$. Now it suffices to observe the sets $S^1 \cap U$ with $U \subset \mathbb R^2$ open are precisely the open sets in relative topology on $S^1$.
Yes, $f(\Bbb R)=S^1$ is closed in $\Bbb R^2$, but it is not relevant for $\tau \subset G$. The statement "Let $s\in S^1$.$\forall r>0,$ we have $B(s,r)\nsubseteq S^1$" is also correct, but it does not have any value for your proof. You can omit it.
To show that $G \subset \tau$, let us consider the continuous surjection $\bar f : \mathbb R \to S^1, \bar f(t) = f(t) = (\cos t, \sin t)$. Here $S^1$ is endowed with $\tau$. For $A \subset S^1$ we have $A = f(f^{-1}(A)) = \bar f(f^{-1}(A))$.
If $A \in G$, then $f^{-1}(A)$ is open and can be written as the union of open intervals $(a,b)$. We claim that all $\bar f((a,b))$ are open in $S^1$; this will prove that $A$ is the union of open subsets of $S^1$, hence is itself open in $S^1$.
The claim is trivial for $b - a > 2\pi$ because then $\bar f((a,b)) = S^1$. For $b - a \le 2\pi$ the closed interval $[a,a+2\pi]$ contains $(a,b)$. The set $K = [a,a+2\pi] \setminus (a,b)$ is compact, hence $\bar f(K) \subset S^1$ is compact, thus closed in $S^1$. Therefore $S^1 \setminus \bar f(K)$ is open in $S^1$. We have $S^1 = \bar f([a,a+2\pi]) = \bar f(K \cup (a,b)) = \bar f(K) \cup \bar f((a,b))$. But $K$ and $(a,b)$ are disjoint, thus $s \in K$ and $t \in (a,b)$ cannot have the same image under $\bar f$ (note that the only two distinct points in $[a,a+2\pi]$ having the same image under $\bar f$ are $a$ and $a+2\pi$ which are both contained in $K$). We conclude that $\bar f(K)$ and $\bar f((a,b))$ are disjoint, hence $\bar f((a,b)) = S^1 \setminus \bar f(K)$ which is open in $S^1$.