let $X$ be a topological space and $\sim$ equivalence relation on $X$. $\pi:X\to X/\sim, x\mapsto [x]$.
A subset $U\subseteq X/\sim$ is open in the quotient topology, iff $\pi^{-1}(U)$ is open in $X$.
This is the finest topology $\tau$ on $X/\sim$, where $\pi$ is continuous.
Therefore the topology with the most open sets.
Is my proof correct?
Suppose $\tau$ is not the finest topology on $X/\sim$. Then there is a set $U\notin\tau$ with $\pi^{-1}(U)$ is open in $X$. But then $U$ has to be in the quotient topology $\tau$, which gives us the contradiction.
Thanks in advance.