Simple question: Suppose $f:X\to Y$ is a nonsurjective map between X,Y topological spaces where Y has the discrete topology. Then there is an $y\in Y$ such that its preimage does not exist; i.e. $f^{-1}(y)$ does not exist. Since Y has the discrete topology, y is closed in Y. Suppose we want f to be continuous, then we want that the preimage of every closed set of Y to be closed in X. But although y is closed, $f^{-1}(y)$ does not exist. Then f is never continuous in this case?
In general, for $f:X\to Y$ nonsurjective map between X,Y topological spaces, how do we think about preimage of open sets of Y when some elements of X are not mapped to elements contained in a open set of Y?
If $y$ is not in the range of $f$ then $f^{-1}(y)$ does exist. It is the empty set and empty set is open in any topology.