# f nonsurjective map between two topological spaces; continuity

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?

• $f^{-1}(y)$ always exists. It is the set of all $x \in X$ such that $f(x) = y$. The only thing is that $f^{-1}(y) = \emptyset$ if $y$ is not in the image f $X$. – Paul Frost Jan 14 at 14:59
• @PaulFrost So general question, suppose y is not in the range of $f$ but $f(a)=f(b)=f(c)=z$, is the preimage $f^{−1}(\{y,z\})=\{a,b,c\}$? – metalder9 Jan 14 at 17:56
• Correct! For any $B \subset Y$ the defnition is $f^{-1}(B) = \{ x \in X \mid f(x) \in B \}$. – Paul Frost Jan 14 at 18:00

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.