Question about functions and topology This is a very general question, but one that I have been struggling with. If we say that a function from a topological space X to a topological space Y is ONTO, then does that mean that for each open set A in Y there is an open set V in X such that f(V)=A, or does it just carry the normal definition that we use in elementary function theory(the definition using points). I know this is an extremely elementary question, but it just happens to be so elementary that I cannot find an adequate answer online.
 A: If the function $f\colon X\to Y$ is continuous and onto, then the following statement is true:

for every open set $A$ in $Y$, there exists an open set $V$ in $X$ such that $f(V)=A$.

How do we show this? Let $A$ be open in $Y$. By the continuity of $f$, we know that $V=f^{-1}(A)$ is open in $X$. Since $f$ is onto (or surjective, it's the same concept), then $f(V)=A$.
Indeed, if $x\in V=f^{-1}(A)$, then $f(x)\in A$, by definition. If $y\in A$, then $y=f(x)$ for some $x\in V$, because $f$ is onto. Then $x\in V=f^{-1}(A)$ and so $y=f(x)\in f(V)$.
If the function $f$ is not continuous this can be true or false, depending on $f$. For instance, consider $X=\mathbb{R}$ with the usual topology and $Y=\{0,1\}$ with the discrete topology. Let
$$
f(x)=\begin{cases}
0 & \text{if $x$ is rational}\\[6px]
1 & \text{if $x$ is irrational}
\end{cases}
$$
Then $A=\{0\}$ is open in $Y$, but there exists no open set $V$ in $X$ such that $f(V)=A$, because no subset of the rationals is open in $\mathbb{R}$.
A: It referees only to the normal set theoretic definition of onto map. For example you can have a bijection(hence an onto map), such as the identity set theoretic map, between $\mathbb{R}$ with the trivial topology and $\mathbb{R}$ with the usual topology but it does not satisfy that for each open set in the second space you have an open set in the first one such as its image is the initial open set in the second space. 
