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Let $X$ be a set and define $f : X → X$

Let $A ⊆ X$

The inverse image (or preimage) is defined as: $$f^{-1}[B]=\{x ∈ X|f(x) ∈ B\} ⊆ X$$

Prove the following:

1) $f(f^{-1}[A]) ⊆ A$

2) $f^{-1}[A]=A⇔f(A)⊆A$

3) $f^{-1}[A]=A⇔f^{-1}[A]⊆A$

I understand (1) visually, and can explain. But I'm uncertain how to prove it. I'm uncertain about (2) and (3).

for (1) I've tried to let $y ∈ f(f^{-1}[A])$ and then try to prove that $y$ is also an element in A, but I couldn't quite figure it out.

I'm told you can do it in a similar way for (2) og (3), where I pick an element on one side, and then prove it's also an element of the other side.

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    1) If $y\in f(f^{-1}[A])$ then you know there is $x \in f^{-1}[A]$ such that $f(x)=y$. But since $x$ lies in preimage of $A$, then what can you tell about $f(x)$? – 2017-02-21
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    I believe you are leading me on to $f(x) = y ∈ A$ ? If that is true, then I think my question is if you could expand on how I know that $f(x)=y$ ? I figure it comes from the definition, but in my mind I see it as $f(x) ∈ y$. I am clearly not understanding it correctly. – 2017-02-21
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    $y\in f(A)$ for some set $A$ in domain of $f$ means that there is some $a\in A$ such that $f(a)=y$. Try to think of it in terms of ordinary functions, if that helps you. Take $f(x)=x^2$ defined on whole real line, then $4\in f(\mathbb{R})$ because $f(2)=4$ for instance. But $-1\notin f(\mathbb{R})$, because there cannot be any real such that $f(x)=x^2=-1$. And by the way $f(x)\in y$ doesn't make any sense, because $y$ is not a set here – 2017-02-21
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    Thank you! I understand your example with a set A... Give me a min or two to see, if I now understand (1) – 2017-02-21
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    Thank you! I believe I got it now. If i let $y ∈ f(f^{-1}[A]) → x ∈ f^{-1}[A]|f(x)=y→f(x)=y ∈ A$. Do you have any suggestions for (2) and (3), or should I be able to work them out using the same basic idea from (1)? – 2017-02-21
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    That looks fine to me. I think (2) and (3) are false. There are some very easy counterexamples for $\impliedby$ directions for both statements. They may become true if some more conditions are imposed on $f$, such as injectivity or surjectivity etc. But for general $f$ they are not true. Only forward direction holds – 2017-02-21
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    I'm told It's correct. the "=>" direction follows from (1), correct? Seem fairly straight forward when I think about them. "<=" or backwards is the tricky one like you said.. Anyways thank you so much for the help, I think I can figure (2) and (3) out now if I look long enough at it. – 2017-02-21
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    Actually, (3) seems to be correct; I was missing something. But for (2), you can have: Let's take $X=\{1,2,3,4\}$. Take $A=\{1,2,3\}$, and define $f:X\to X$ by $f(n)=n$ for $n=1,2,3$ and $f(4)=1$. Then $f(A)=A$ (so $f(A)\subset A$ in particular), but $f^{-1}[A]=X \neq A$. – 2017-02-21
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    (2) is incorrect? Give me a few minutes to look it over and see if I understand it. You say you get (3) as correct. Care to write it out, as (2)? As (3) is the hardest for me to understand. – 2017-02-21
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    At (2), can't I do like this: $f^{-1}[A]=A→f^{-1}[A]⊆A$ (this is obvious). And $f^{-1}[A]=A→f(A)=f(f^{-1}[A])⊆A$ And therefore it is correct? – 2017-02-21
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    You only proved $\implies$ direction, not the other one. Also for (3) on the same setting as in (2), now take $f(3)=3$, $f(4)=4$ and $A=\{2,3,4\}$, $f(1)=f(2)=1$. Then $f^{-1}[A]=\{3,4\}\subset A$ which is not equal to $A$. So (3) is also incorrect for $\impliedby$ direction. – 2017-02-21
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    Oh yes, that is correct. I will have to look at this tomorrow, but thank you (again) so much for the help. – 2017-02-21

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