TITLE: topology - Quotient topology
QUESTION [6 upvotes]: Let $k:X \to Y$ be an onto map.How to prove that the quotient topology on $Y$ induced by $k$ is the largest topology relative to  which $k$ is continuous.

REPLY [10 votes]: I’m going to assume that you’ve defined the quotient topology $\tau$ on $Y$ as follows: a set $U\subseteq Y$ is open iff $k^{-1}[U]$ is open in $X$. It’s an immediate consequence of this definition that the topology $\tau$ on $Y$ makes $k$ continuous, so what you have to show is that if $\tau'$ is another topology on $Y$ making $k$ continuous, and $\tau\subseteq\tau'$, then $\tau=\tau'$. In other words, you must show that it’s impossible to have $\tau'\supsetneqq\tau$: there is no topology on $Y$ that is strictly stronger than $\tau$ and makes $k$ continuous.
Here’s a large hint to get you started. Suppose that $\tau\subseteq\tau'$, where $\tau'$ is a topology on $Y$ making $k$ continuous. If $\tau'\ne\tau$, there is a set $U\in\tau'\setminus\tau$. By hypothesis $k$ is continuous as a map from $X$ to $\langle Y,\tau'\rangle$, so $k^{-1}[U]$ is open in $X$. Now apply the definition of the quotient topology $\tau$ to get a contradiction.