Let $f:X \rightarrow Y$ and let $A \subset X$ and $B \subset Y$. Then

1) $A \subset f^{-1} \circ f(A) $

2) $f \circ f^{-1} (B) \subset B$

I'm a little stuck mainly by the notation in this problem. I understand the proof for compositions but I think where I'm getting stuck on is using the composition of the function with its inverse.

  • $\begingroup$ This has nothing to do with topology, per se. It's basic set theory. (I edited tags accordingly.) $\endgroup$ – BrianO Mar 8 '16 at 1:53
  • $\begingroup$ Only used the topology tag as it was in a topology text. I'm just starting the subject, my mistake. $\endgroup$ – Chris Millett Mar 8 '16 at 1:55
  • $\begingroup$ No problem. But just so you know :) Notice that the proof uses no topological property of $A$ or $B$ — they're just sets — and $f$ is just a function, not assumed to be continuous. $\endgroup$ – BrianO Mar 8 '16 at 2:27

First notice that $f^{-1}\circ f(A)=f^{-1}(f(A))$ and $f\circ f^{-1}(B)=f(f^{-1}(B))$ by definition.

So for $1)$ you want to show that for any $x\in A$ we have $x\in f^{-1}(f(A))$. And for $2)$ you want to show that for any $x\in f(f^{-1}(B))$ we have $x\in B$.

Lets look at $1)$ together and maybe you can do $2)$ by using similar reasoning. Let $x\in A$. Then $f(x)\in f(A)$. Looking at the set $f^{-1}(f(A))$, which by definition is $$f^{-1}(f(A))=\{x\in X:f(x)\in f(A)\},$$ we conclude that indeed $x\in f^{-1}(f(A))$. So our conclusion is that $A\subset f^{-1}(f(A))$.

  • $\begingroup$ That was great thank you! I think this is the best way to do it. Thanks for not holding my hand through it $\endgroup$ – Chris Millett Mar 8 '16 at 1:49
  • $\begingroup$ @ChrisMillett. Glad to help! $\endgroup$ – T. Eskin Mar 8 '16 at 1:53

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