If X is a space and A is a set and if $p : X \rightarrow A$ is a surjective map, then there exists exactly one topology $\tau$ on A relative to p which is a quotient map ; It is called the quotient topology induced by p.

The topology $\tau$ is ofcourse defined by letting it consist of those subset U such that $p^{-1}(U)$ is open in X. So I thought about proving that the quotient top is unique by proving it is the finest top on Y that make f continous, but why is it the finest top ?

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    $\begingroup$ It’s unique because it’s defined to be a specific topology on $A$. $\endgroup$ – Brian M. Scott Oct 26 '15 at 3:21
  • $\begingroup$ but couldn't we have more than one topology on A that makes f continuous ? $\endgroup$ – Illustionist Oct 26 '15 at 3:23
  • $\begingroup$ @Illustionist: Of course there can more than one topology that makes $f$ continuous. But there's exactly one that is the finest out of all such topologies. $\endgroup$ – Zev Chonoles Oct 26 '15 at 3:26
  • $\begingroup$ but why is quotient topology defined above is the finest one ? $\endgroup$ – Illustionist Oct 26 '15 at 3:27
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    $\begingroup$ Your definition is a bit weird. I have usually heard that the quotient topology on $A$ is the finest topology on $A$ such that $p$ is continuous. (It will be the finest one because, since $A$ is the codomain, it is "harder" for $p$ to be continuous if $A$ has a finer topology. Thus the continuity "buys us more" the finer we make the topology of the codomain. We want it to buy it as much as possible, so we take the finest one.) $\endgroup$ – Ian Oct 26 '15 at 3:27

let $\rho$ be the collection of open sets defining the topology of the space X. define the following collection of subsets of A:

$$ \sigma_p = \{U \in \mathfrak{P}(A)|p^{-1}(U)\in \rho \} $$

suppose $\tau$ is any other topology on A making $p$ continuous. then for $V \in \tau$ we have $p^{-1}(V) \in \rho$ hence $V \in \sigma_p$. since V is arbitrary this means that $\tau \subseteq \sigma_p$


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