Equivalence of pretopological continuity to customary continuity (a direct proof)

Question

Can you provide a direct (not based on the neighborhood definition of pretopologies) proof for pretopological spaces (expressed as closure operators) that a function $f$ from a topological space $\mu$ to a topological space $\nu$ is continuous (in the customary topology sense) that is preimage of every open set is open if and only if it is continuous in the pretopology sense that is when $f[\operatorname{cl}_{\mu}A] \subseteq\operatorname{cl}_{\nu} f[A]$ for every subset $A$ of space $\mu$?

Answer 1

Given topological spaces $X,Y$ and a function $f: X \to Y$, I'll show that the two forms of continuity are equivalent. Note that I'll be using the overline $\overline{\phantom{A}}$ to denote the closure operator in both topological spaces.

• Assume that $f : X \to Y$ is continuous "in the customary topological sense". Given $A \subseteq X$, note that $\overline{f[A]}$ is closed in $Y$, so $Y \setminus \overline{f[A]}$ is open in $Y$, and therefore $f^{-1} [ Y \setminus \overline{f[A]}] = X \setminus f^{-1} [ \overline{ f[A] } ]$ is open in $X$. It follows that $f^{-1} [ \overline{ f[A] } ]$ is closed in $X$. Note, too, that $$A \subseteq f^{-1} [ f [ A ] ] \subseteq f^{-1} [ \overline{ f[A] } ],$$ and so it follows that $$\overline{A} \subseteq f^{-1} [ \overline{ f[A] } ].$$ We now have that $$f [ \overline{A} ] \subseteq f [ f^{-1} [ \overline{ f[A] } ] ] \subseteq \overline{f[A]},$$ as desired.

• Assume that $f : X \to Y$ is continuous "in the pretopological sense". Given an open $V \subseteq Y$, we want to show that $f^{-1} [ V ]$ is open, or, equivalently that $X \setminus f^{-1} [ V ] = f^{-1} [ Y \setminus V ]$ is closed. That is, we want to show that $\overline{ f^{-1} [ Y \setminus V ] } = f^{-1} [ Y \setminus V ]$. By properties of closure, we already know that $f^{-1} [ Y \setminus V ] \subseteq \overline{ f^{-1} [ Y \setminus V ] }$, and so we need only show the reverse inclusion.

  By pretopological continuity, $f[ \overline{ f^{-1} [ Y \setminus V ] } ] \subseteq \overline{ f[ f^{-1} [ Y \setminus V ] ] } \subseteq \overline{ Y \setminus V} = Y \setminus V$ (recall that $Y \setminus V$ is closed). Therefore $$f^{-1} [ Y \setminus V ] \supseteq f^{-1} [ f [ \overline{ f^{-1} [ Y \setminus V ] } ] ] \supseteq \overline{ f^{-1} [ Y \setminus V ] },$$ as desired.

(Note that in the above I am making use of such facts as $f^{-1} [ f [ A ] ] \supseteq A$ and $f [ f^{-1} [ B ] ] \subseteq B$.)