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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.

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    But what *is* the quotient topology, according to you? It might help if you spelled out all the definitions in detail. What is the quotient topology induced by k and what is the largest topology with respect to which k is continuous? Write them down and then it might be possible to give an answer different from "by definition".2012-10-29

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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.

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    @ccc: Recall that a set $V\subseteq Y$ is in $\tau$ iff $k^{-1}[V]$ is open in $X$, by the definition of the quotient topology $\tau$. And $k^{-1}[U]$ must be open in $X$, since $k$ is continuous with respect to $\tau'$. But then $U\in\tau$, contrary to our choice of $U$.2012-11-08