Characterising open sets in a topological space Suppose we have topological spaces $X,Y$ and a continuous map $f:X\rightarrow Y$.
Is it true that any open set in $X$ can be written in the form $f^{-1}(U)$ for some open set $U \subset Y$?
The reason I ask is that my professor used this fact in proving that a particular map is open; yet it seems to me that this fact is equivalent to $f$ being open. Am I missing something? Is it true when $X$ is a topological subspace of $Y$?
 A: This is not true.  Consider $X = \lbrace a,b, c\rbrace$ and $Y = \lbrace y \rbrace$ both with the discreet topology.  Let $f\colon X\to Y$ be defined by $f(x) =y$ for all $x\in X.$  Then $f$ is clearly continuous, but $f^{-1}(U) = X,$ for all nonempty $U\subseteq Y$, and there is no way to write, for example, $\lbrace a \rbrace$ as the preimage of a set in $Y$ under $f$.
As for the possibility that this be true for $X\subset Y$, consider $X = \lbrace x,y\rbrace$ and $Y = \lbrace x,y,z\rbrace$, both again with the discreet topology.  Let $f\colon X\to Y$ be defined by $f(x) = y$ for all $x\in X.$  Again $f$ is continuous.  Here we can characterize $f^{-1}(U)$ for sets $U\subseteq Y$ by whether or not they contain $y$.  If $y\in U$, then $f^{-1}(U) = X$.  If $y\not\in U$, then $f^{-1}(U)=\emptyset$.  Therefore we can't, for example, find a set $U\subseteq Y$ such that $f^{-1}(U) = \lbrace x \rbrace,$ and so this is not true here either.
A: Easily seen to be false . Suppose $f(p)=f(q)$ and there is an open set $U$ in $X$ with $p\in U$ and $q\not \in U.$ If $V\subset Y$ and $p\in f^{-1}V$ then $f(p)\in V$, so $$f^{-1}V\supset f^{-1}\{f(p)\}=f^{-1}\{f(q)\}\supset \{q\}$$ so $f^{-1}V\ne U.$
Even if $f$ is a continuous injection it can fail. For example $X=[0,1),\; Y=\{z\in \mathbb C: |z|=1\}$ and $f(x)=e^{ix}.$ Then $[0,1/2)$ is open in $X$ but $f^{-1}V=[0,1/2)\iff V=\{e^{ix}: x\in [0,1/2)\}$ which is not open in $Y. $ 
Were there additional conditions on the function that the professor was talking about?
