Why does there exist a unique quotient topology that makes a given surjective map a quotient map?

I'm reading Munkres on the quotient topology, pg. 138

It says that:

If $X$ is a topological space and $A$ is a set and if $p: X \to A$ is a surjective map, then there exists exactly one topology on $A$ relative to which $p$ is a quotient map.

It then goes on to say that this topology is defined by letting the open sets be those subsets $U$ of $A$ such that $p^{-1}(U)$ is open in $X$.

I understand why this is a topology, and I understand why this makes the $p$ map a quotient map. I just don't understand why this is the only topology to make $p$ a quotient map. I understand that it may be a minimal topology to make $p$ a quotient map.

It was my understanding that being a quotient map is a rather weak condition compared to others such as being an open map (when surjectivity is assumed). So couldn't we have defined a topology on $A$ that makes $p$ into an open map, whereby it would then be a quotient map? This topology would be finer than the one given above and could be defined by having the open sets $U$ be those where there exists an open set $V$ in $X$ such that $p(V) = U$. Is there any substantive difference here or would they essentially be the same topologies?

• Isn't quotient map defined as a map where $U$ is open iff $p^{-1}(U)$ is open? – Stefan Hamcke Oct 3 '13 at 2:26
• @StefanH. yes, thank you. I just remembered that open map implies quotient map only when the map is both surjective and continuous. I had forgotten the continuous part, which explains my confusion. – user96787 Oct 3 '13 at 2:32

The final topology on $A$ with respect to $p$ is the finest topology for which $p$ is continuous. Every subset of $A$ that may be open (whose preimage under $p$ is open) is open in the final topology. Since a quotient map is among other things continuous, you cannot take a strictly finer topology on $A$ and have $p$ still be a quotient map.