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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?

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usereb
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    Define $V(x)$ for us, please. The set of “neighbourhoods” of $x$, I think. But what then is a neighbourhood of $x$? And continuous is defined as local continuity at each point? Many take 2. to be the definition. – Henno Brandsma Jan 15 '18 at 18:22
  • @HennoBrandsma Since I have indicated $V(f(x))$ is nbd of $f(x)$ I have not indicated $V(x)$ is nbd of $x$ again. I know b and c are generally being used as definition bu it was my final exam question. It is not written what then is a nbd of $x$, also I couldn’t see why in some websites – usereb Jan 15 '18 at 18:38

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$f^{-1}(O)=\{x\in X\ |\ f(x)\in O\}$ 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).

  1. If $x \in f^{-1}(O)$, then $f(x) \in O$ by definition. $O$ is open, so $O\in\tau_2$.

  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 $f^{-1}(O)\in V(x)$.

  3. Again by a property of the inverse image: $f^{-1}(Y\setminus F)=f^{-1}(Y)\setminus f^{-1}( F)=X\setminus f^{-1}(F)$.

The Phenotype
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    Thank you very very much. I have been considering $f$ as a function. Your answer is very beneficial for me – usereb Jan 16 '18 at 09:28
  • Why -1? I ve given +1. Is there any mistake? It looks true? – usereb Jan 25 '18 at 08:50
  • @esrabasar No need to worry. Someone downvoted all my questions and most of my answers, so the downvote is not due to the content, else he would certainly write a comment about what would be wrong with my answer. – The Phenotype Jan 25 '18 at 11:48
  • It is really weird. You can contact with stackexchange. Have a nice day :) – usereb Jan 25 '18 at 14:25