Let $f:(X,\tau_1)\to (Y,\tau_2)$ be a function. TFAE :
a) $f$ is continuous on $X$
b) $\forall O \in \tau_2 $ $f^{-1}(O) \in \tau_1$
c) $\forall F$ closed in $Y$, $f^{-1}(F)$ is closed in $X$
In many proofs I've seen that in a $\Rightarrow$b part : $x \in f^{-1}(O) \Rightarrow$ $f(x) \in O \in \tau_2$
1) How can we say $x \in f^{-1}(O) \Rightarrow$ $f(x) \in O \in \tau_2$ above? We don't know $f$ is surjective or not?
So $O \in V(f(x))$ -nbd of $f(x)$- , $f^{-1}(O) \in V(x)$
2) It is clear $O \in V(f(x))$ but I didn't understand how we said that $f^{-1}(O) \in V(x)$
And in b $\Rightarrow$c and c $\Rightarrow$a parts :
$F^C=O$, $f^{-1}(O)=f^{-1}(Y\setminus F)=$ $X\setminus f^{-1}(F)$
3) How did we say $f^{-1}(Y\setminus F)=$ $X\setminus f^{-1}(F)$, again we dont know $f$ is surjective or not. It is valid if $f$ is surjective like above one, isn't it?
Could someone illuminate me please?
$f^{-1}(O)=\{x\in X\ |\ f(x)\in O\}$ means the inverse image of $O$ under $f$, not the inverse of $f$ evaluated in an element (note that $O$ is a subset of $X$, not an element of $X$). So you don't need to know anything about $f$ (except for it being a function with $O$ a subset of its domain).
If $x \in f^{-1}(O)$, then $f(x) \in O$ by definition. $O$ is open, so $O\in\tau_2$.
$O \in V(f(x))$ implies $f^{-1}(O) \in V(x)$, because $f(x)\in O$ and $O$ is obviously a neighborhood (of $f(x)$), so by definition of continuity of $f$ we get that there exists a neighborhood $U$ of $x$ such that $f(U)\subseteq O$, which is equivalent to writing $U\subseteq f^{-1}(O)$, so $f^{-1}(O)\in V(x)$.
Again by a property of the inverse image: $f^{-1}(Y\setminus F)=f^{-1}(Y)\setminus f^{-1}( F)=X\setminus f^{-1}(F)$.