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Sorry for not using the correct notation yet, I just joined:)

The previous question had a typo in it, it's been corrected now.Here is my now conclusion: I want to prove the statement is true and thought of element chasing to prove it. Is the following somewhat correct?

Take $m \in f(f^-1(A))$

Since A is a subset of Y, $(f^-1(A)) \in x$

So $m \in f(x)$

Thus $m \in y$. I'm not sure if this is correct, as it doesn't seem like I've proved that $f(f^-1(A))$ is a subset of A, but rather that it leads to any input being sent to Y.

Thank you!

Layla16
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    If $A \subseteq X$ then $f^{-1}(A)$ is not defined. Do you mean $A \subseteq Y$? – Jair Taylor Oct 26 '23 at 18:17
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    Is the question not asking about a function when its inverse is given as input? @AnotherUser – Layla16 Oct 26 '23 at 18:18
  • You're right that $f^{-1}(A)$ generally doesn't make sense if $f : X \rightarrow Y$ and $A \subseteq X$. Was there perhaps a typo, and it should be $A \subseteq Y$? – Sambo Oct 26 '23 at 18:18
  • @JairTaylor This was the question which I'm trying to prove or disprove, if it helps the question states to either prove the statement or come up with an example to prove it doesn't hold. – Layla16 Oct 26 '23 at 18:20
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    The notation $f^{-1}(A)$ denotes the preimage (or inverse image) of $A$. – Another User Oct 26 '23 at 18:20
  • For clarification: there was a typo in the question, it was supposed to be ⊆. Sorry! – Layla16 Oct 26 '23 at 19:52
  • @Layla16: As Another User mentions, $f^{-1}(A)$ refer to the preimage of $A$ under $f$, which is defined as ${x\in X:f(x)\in A}$. This notation is potentially ambiguous in the situation that the inverse function $f^{-1}$ exists, and $A$ is a both a subset of $Y$, and a member of $Y$, but this situation rarely happens in practice... – Joe Oct 26 '23 at 19:56
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    Take $y \in f(f^{-1} (A)).$ Then there exists $x \in f^{-1} (A)$ such that $y = f(x).$ But since $x \in f^{-1} (A)$ it follows that $f(x) \in A$ and hence $y \in A.$ So $f(f^{-1} (A)) \subseteq A.$ This completes the proof. Pretty straightforward! – Akiro Kurosawa Oct 26 '23 at 20:16

3 Answers3

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I will give an informal but intuitive reason as to why this is true. Take any element $y$ in $A$. If you take the preimage of this point one of two things will happen:

  1. You get a point $x \in X$ such that $f(x) = y$.
  2. You get the empty set. This happens when $y$ is not in the range of $f$ (i.e. $f$ is not necessarily surjective). Think about what happens when you take the preimage of $-1$ under the map $f: \mathbb{R} \rightarrow \mathbb{R}$, $f(x) = x^2$.

Now map everything back to $Y$ by using $f$. If (1) occurred, you will get $y$ back by definition. If (2) happened then you will get nothing since the image of the empty set is again the empty set. Hence you should expect that $f(f^{-1}(A))$ is at most $A$, or written in a more mathematical way, $f(f^{-1}(A)) \subset A$. Again the key point here is $f$ may not be surjective.

To make this all into a legitimate proof your idea of element chasing is the correct one. If you can show that any $y \in f(f^{-1}(A))$ must also be in $A$ then you are done since $y$ was chosen arbitrarily. Akiro Kurosawa's comment explains this nicely.

CBBAM
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You have numerous notational errors in your question - forgetting capitalization (which matters in math) and confusing $\in$ with $\subset$, for example, since $f^{-1}(A)$ is a set. It also appears you may be confusing the preimage of a set with the inverse function applied to a set (which may not exist here since you never specify that $f$ is invertible).

However, it appears your reasoning is incorrect for the reason you state - you conclude that $f(f^{-1}(A)) \subset X$, not $f(f^{-1}(A)) \subset A$.

However, it is true that $f(f^{-1}(A)) \subset A$, more or less by definition: $f^{-1}(A)$ is defined as the set of $x \in X$ such that $f(x) \in A$. So $f(x) \in A$ for all $x \in f^{-1}(A)$, thus $f(f^{-1}(a)) \subset A$.

MartianInvader
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If $a\in f(X)$ then $ff^{-1}(a)=a$. So, $$ff^{-1}(A)=A\bigcap f(X)\subset A$$

Bob Dobbs
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