Revisit the following discussion:
Prove that the inverse image of an open set is open
Obviously, the above discussion is based on Euclidean space (which is also a metric space, so the proof is based on the open ball). Can we say the following:
Let $X,Y$ be any two topological spaces,
$f: X \rightarrow Y$ be a continuous function. The inverse image of an open set is open under $f$.
Can this famous theorem apply to any topological space? for example, Zariski space?
Let $X,Y$ be top. spaces and $f:X \to Y$ a function. $f$ is called continuous at $x_0 \in X$ , if for every neighborhood $V$ of $f(x_0)$ there is a neighborhood $U$ of $x_0$ such that $f(U) \subseteq V$.
$f$ is called continuous on $X$ ,if $f$ is continuous at every $x \in X$.
Then we have:
$f$ is continuous on $X \iff$ for each open set $B \subset Y$ the inverse image $f^{-1}(B)$ is open (in $X$).
Actually, that's the usual way of defining the meaning of “continuous function” in Topology textbooks. So, it's not a theorem at all; it's a definition.
Of course, you can also say that a function $f\colon X\longrightarrow Y$ is continuous if, for each $x\in X$ and each neighborhood $V$ of $f(x)$, $f^{-1}(V)$ is a neighborhood of $x$. If you adopt this as the definition of “continuous function” then, yes, your statement becomes a theorem which has to be proved. It's quite easy.
And, yes, in particular this applies to the Zarisky topology.
