Quotient topology, finest topology

Question

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.

Answer 1

The easiest way to reason, I think is:

Suppose that $\mathscr{T}$ is any topology on $X/ {\sim} $ that makes $\pi$ continuous. Let $U \in \mathscr{T}$, then by continuity of $\pi$ gives us that $\pi^{-1}[U]$ is open in $X$. But then the definition of the quotient topology says that $U \in \mathscr{T}_{\text{quot}}$. So $\mathscr{T} \subseteq \mathscr{T}_{\text{quot}}$.

This shows that any topology that makes $\pi$ continuous is a subset of the quotient topology. And clearly the quotient topology is one of the topologies that makes $\pi$ continuous. This shows that $\mathscr{T}_{\text{quot}}$ is the maximal (i.e. finest) topology that makes $\pi$ continuous.

Answer 2

Your reasoning is correct. However the quotient topology is usually defined as the finest topology. Then it is to be proved that $U$ is open by quotient topology iff $\pi^{-1}(U)$ is open within the domain space of $\pi$.