Continuity vs. strong continuity

Question

Why is this condition stronger than continuity? By definition, a map is continuous iff the preimage of an open set is open. That's precisely the condition in the cited definition, isn't it?

Answer 1

The distinction is that normal continuity of a map $f: X \to Y$ cannot be used to determine that a subset $U$ of $Y$ is open. Continuity gives the implication that $f^{-1}(U)$ is open provided $U$ is open, and being a quotient map means that you can additionally infer that $U$ is open provided that $f^{-1}(U)$ is open.

As an example, consider the map $f_{0}: \Bbb R \to \Bbb R$ sending all points to 0. Note that constant maps are always continuous. Now $f^{-1}(\{0\})$ is open, being all of $\Bbb R$. However $\{0\}$ is not open, demonstrating that $f_0$ is continuous but does not satisfy this stricter condition.

edit: Thanks to Harry Altman in the comments for noting that I was ignoring the surjectivity requirement. For a surjective example, consider the identity map $X_{\textrm{discrete}}\to X_{\textrm{indiscrete}}$ where the $X_{\textrm{discrete}}$ and $X_{\textrm{indiscrete}}$ is the discrete and indeiscrete topologies on an arbitrary set $X$. Here $f$ is continuous (follows from either the domain being discrete or the codomain being indiscrete) but for any nonempty proper $U \subset X$ we have that $U$ is not open in $X_{\textrm{indiscrete}}$ but $f^{-1}(U)$ is open in $X_{\textrm{discrete}}$.