Understanding Saturated Sets in Quotient Maps

[hideelo]

I am reading munkres topolgy and I am struggling with understanding the following sentence:

"We say that a subset C of X is saturated (with respect to the surjective map p:X→Y) if C contains every set p^-1({y}) that it intersects"

if you have the second edition its in chapter 2 section 22 (page 137)

It's not that I have questions on it I just can't seem to make heads or tails of that sentence.

any help would be appreciated

 

[Jorriss]

"We say that a subset C of X is saturated (with respect to the surjective map p:X→Y) if C contains every set p^-1({y}) that it intersects"

All it means is that if C intersect p^-1({y}) is nonempty, then C actually contains all of p^-1({y}). So if C is saturated and p^-1({y}) has say, two elements, it is not possible that only one of those elements is in C.

This is equivalent to "C is a saturated subset of X if C is the preimage of some subset of Y".

 

[hideelo]

thanks

do we make any demands on the subset of Y, need it be open, closed... or will any subset do?

 

[Jorriss]

thanks

do we make any demands on the subset of Y, need it be open, closed... or will any subset do?

Any subset will do.

 

[micromass]

thanks

do we make any demands on the subset of Y, need it be open, closed... or will any subset do?

Another equivalent form is that $C$ is saturated if and only if $C=p^{-1}(p(C))$. So the exact form of the subset of $Y$ is $p(C)$.

That said, I prefer dealing with quotient mappings in terms of equivalence relations. Thus a quotient map $p:X\rightarrow Y$ induces a equivalence relation on $X$ given by $x\sim x^\prime$ if and only iff $p(x) = p(x^\prime)$. The set of all equivalence classes can then be identified wuth $Y$.

In that form, we can give new equivalent forms of saturated sets. One such form is to say that a set $C$ is saturated iff it is the union of equivalence classes. Another form is to say that $C$ is saturated if for any $x\in C$ and $y\in C$ such that $x\sim y$ holds that $y\in C$.

All of these forms are easily seen to be equivalent, but sometimes one equivalent form might give more insight than another one.