# The name for the quotient property.

We call a surjective $f:X\rightarrow Y$ a quotient mapping if it satisfies, for every $U\subset Y$

(continuity, continuous) $U$ is open $\Rightarrow$ $f^{-1}(U)$ is open

and

(???) $f^{-1}(U)$ is open $\Rightarrow$ $U$ is open.

I was wondering if there is the name for the second property itself of the definition, or should I just call it "the other" property? And if a function satisfies this property only, how should I call it?

## 2 Answers

Normally we only consider the continuity as given, and then the property that ($f^{-1}[U]$ open implies $U$ is open) is what makes $f$ a quotient map (by definition!).

So a continuous onto $f: X \rightarrow Y$ is defined to be a quotient map, iff for all $U \subset Y$: $f^{-1}[U]$ open in $X$ implies $U$ open.

Or equivalently $f: X \rightarrow Y$ (onto function) is a quotient map iff for all $U \subset Y$: $U$ open iff $f^{-1}[U]$ open.

So your "other" is just the typical property for being a quotient map. Normally this property is not considered without continuity of $f$ as well.

• You last sentence "Normally this property is not considered without continuity of $f$ as well." is something what I am looking for in this question. So, basically, what you are saying, there is no name for it, and if a function satisfies this property only, I just state the whole property without being able to just name it. – Vadim Oct 21 '14 at 18:45
• @Vadim At least I haven't seen it used that way. But of course there might very well be papers written on it that I haven't read. You might try MathOverflow as well, people there might know it under some name. – Henno Brandsma Oct 21 '14 at 20:16

A more general name for this kind of topology is "final." This involves a change of perspective: we imagine the topology on $Y$ is not already given, but that it's going to be defined by the quotient map $q$. We say that a quotient space has the final topology induced by $q$. "Final" here means that the topology is the finest possible one that makes $q$ continuous, that is, we put in as many open sets as possible. What's a possible open set? Well, if $q$ is continuous and $U$ is going to be open, then we know we must have $q^{-1}(U)$ open. It's quick to check that if all such $U$ are declared to be open, $Y$ becomes a topological space admitting a continuous surjection $q$ from $X$, which justifies calling this the final topology.

We can define final topologies induced by more than one map, which is where we begin to generalize. For instance, any object defined by gluing charts together, in particular, vector bundles and manifolds, have the final topology induced by all their charts. This is generally an uncountable collection, so it's a very different situation than a quotient map. For another example, any open map induces the final topology on its image, so that a surjective open map is automatically a quotient map; similarly for a surjective closed map (which is already open.) But this implication is not reversible-there are quotient maps that are neither open nor closed!

• I'm sorry, I don't see how this even closely answers the question. I know all this, but the question is not about quotient/final topologies. The question is about the name for the second property, in general, and a name for the function that satisfies the second property only, in particular. It is very possible, as @Henno already suggested that there is no name for it. – Vadim Oct 21 '14 at 18:39
• Ah, sorry, I didn't catch that you wanted a name for a not-necessarily continuous map that satisfies the second property. I agree that there's probably not a name for it in itself, because there are probably no natural examples of such maps that aren't continuous. – Kevin Carlson Oct 21 '14 at 22:08
• That's OK, I should have probably be more clear. Such functions are rarely considered out of the context where they are continuous (so quotient) at the same time. But this does not mean they are rare. Take any quotient map, relax the topology of the image, and you get a continuous non-quotient map, make the topology finer -- and you have a non-continuous function that still satisfies the second property. – Vadim Oct 22 '14 at 3:33
• Well, agreed that it's easy to construct such a map, but that example doesn't look obviously natural to me. Maybe it comes up in functional analysis or something when one works at several levels of the lattice of topologies simultaneously? Anyway, that's to one side. – Kevin Carlson Oct 22 '14 at 16:17