Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm having a bit of trouble with an example problem in the topology book I'm reading. It's problem #11 (pp 104) of the "Solved Problems" section of Chapter 7, of the Schaum's Outline for "General Topology" (Seymour Lipschutz, ISBN 0070379882).

In the proof outlined in the example, the following statement is made at an intermediate step of the proof: $$\overline{f[A]} = f[f^{-1}[\overline{f[A]}]]$$ And again later as part of the same proof (in the other direction of the iff), a similar statement is made: $$\overline{f[f^{-1}[F]]}=\bar{F}$$

I can see how these statements would follow if the $f$ were one-to-one. But I can't see how that would be true in-general. Is there something about the closure of a set, or a closed set in-general, which makes these statements true?

Theorem 2.8 (pp 21) in the same book states that $A \subset f^{-1} \circ f[A]$ and $B \supset f \circ f^{-1}[B]$, but I can't see when those become equalities (other than when $f$ is one-to-one).

Without making the assumptions in the above statements, I am able to come to the same conclusion as the proof does, by a different route. But I'm still curious whether there is something elementary which I am not quite understanding.

If more context is necessary, I can update this post with the entire example, and my alternate solution to it. But I just wanted to get the initial question out there, in case there is a simple answer that doesn't depend on the proof as a whole.

share|improve this question
    
A similar statement is made again in a later example (solved problem #18, pp 106), this time where $N$ is a neighborhood: $$f[f^{-1}[N]]=N$$ –  PaulDeLong May 7 '13 at 16:08

1 Answer 1

up vote 4 down vote accepted

Let $f:X\to Y$, and let $A\subseteq Y$. Then $f\big[f^{-1}[A]\big]=A$ iff $A\subseteq f[X]$. To see this, note that you always have $f\big[f^{-1}[A]\big]\subseteq A$: if $x\in f^{-1}[A]$, then by definition $f(x)\in A$. And if $A\subseteq f[X]$, then for each $a\in A$ there is an $x\in f^{-1}[A]$ such that $f(x)=a$, so $a\in f\big[f^{-1}[A]\big]$.

If $A\subseteq X$, it is of course always true that $f^{-1}\big[f[A]\big]\supseteq A$, with equality if $f$ is injective; beyond that there’s not much to be said in general.

share|improve this answer
    
Apologies for taking so long to upvote this. I've been busy and away from StackExchange for a while, and only just found the time to get back to this. –  PaulDeLong Nov 26 '13 at 18:53
    
@Paul: No problem! (My goodness, that was a while.) –  Brian M. Scott Nov 26 '13 at 19:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.