Why can the quotient topology not be generated by images of the projection map? If $Q$ is a topological space and $\tilde{Q}$ is the quotient space of $Q$ under the equivalence relation $\sim$. Let $\pi$ be the projective map sending $q \in Q$ to its equivalence class $[q] \in \tilde{Q}$. Then by definition the quotient topology of $\tilde{Q}$ consists all sets $U$ where $\pi^{-1}(U)$ is open in $Q$.
I am having some trouble seeing why the above definition is not equivalent to the statement "the quotient topology on $\tilde{Q}$ consists of all open sets $U$ of the form where $U = \pi(V)$ for some open $V$ in $Q$". This definition would imply that $\pi$ is open, but this is not the case as I have seen multiple counterexamples. However, I'm still having some difficulty seeing where the topology generated in this fashion is not the same as the one generated by looking at preimages of sets.
 A: As azif00 comments, the set $\sigma = \{\pi(V) \mid V \subset Q \text{ open}\}$ is not necessarily a topology on $\tilde Q$.
However, we can consider the topology $\tau(\sigma)$ having $\sigma$ as a subbase. Then $\pi : Q \to \tilde Q$ becomes an open map. But if we endow $\tilde Q$ with the quotient toplogy, $\pi$ is in general not an open map which shows that the two topologies are in general distinct. What is the reason for this fact?
The quotient topology $\tau_\pi$ is always a subset of $\sigma$. Its elements are the subsets $U \subset \tilde Q$ for which $\pi^{-1}(U)$ is open in $Q$, thus they have the form $U =\pi(\pi^{-1}(U))$ , i.e. are images of special open subsets of $Q$. But in general not every open $V \subset Q$ has the form $V = \pi^{-1}(U)$, thus in general $\tau_\pi \subsetneqq \sigma \subset \tau(\sigma)$.
Summarizing, $\tau(\sigma)$ is always finer than $\tau_\pi$ and in general even strictly finer.
Remark.
If $f : X \to Y$ is a  function, a subset $S \subset X$ is called saturated if $S = f^{-1}(f(S))$. Note that $S \subset X$ is saturated if and only if it has form $S  = f^{-1}(T)$ for some $T \subset Y$. For the only-if-part take $T = f(S)$. For the if-part observe that $S \subset f^{-1}(f(S))$ for all $S \subset X$ and $f(f^{-1}(T)) \subset T$ for all $T \subset Y$ so that $S \subset f^{-1}(f(S)) = f^{-1}(f(f^{-1}(T))) \subset  f^{-1}(T) = S$ which implies $S = f^{-1}(f(S))$.
Thus the open subsets of $Q$ having the form $\pi^{-1}(U)$ for some $U \subset \tilde Q$ are precisely the open saturated subset of $Q$. Therefore  $U \in \tau_\pi$ if and only if it is the image of an open saturated subset of $Q$. In fact, $U \in \tau_\pi$ means that $\pi^{-1}(U)$ is open in $X$. As we have seen above, $\pi^{-1}(U)$ is saturated and $\pi(\pi^{-1}(U)) = U$ since $\pi$ is surjective; this proves the only-if-part. For the if-part let $U = \pi(V)$ with an open saturated $V \subset Q$. Then $\pi^{-1}(U) = \pi^{-1}(\pi(V)) = V$ is open in $Q$, thus $U \in \tau_\pi$.
