Set theory - Image, preimage I have this assignment.
Let $X$ be a set and $f:X\to X$ a function. Let $A\subseteq X$. Determine whether each of the following statements is true in general, and give a proof of the correct statements. Give a counterexample for each of the statements that is not true in general.
a) $f(f^{-1}(A))\subseteq A$
b) $f(f^{-1}(A))=A$
c) $f^{-1}(A)=A \Leftrightarrow f(A)\subseteq A$ and $f^{-1}(A)\subseteq A$.
Update: For a) and c) I now know what to do. Look at the great answers below.
For b) I can find a counterexample:
Let $X=\{1,2\}$ og $A=\{1\}$. Let $f$ be the function that:
\begin{align*}
f(1)=2\\
f(2)=2
\end{align*}
Then:
\begin{align*}
f(A)=\{2\}\\
f^{-1}(A)=Ø\\
f(f^{-1}(A))=Ø
\end{align*}
It follows that
\begin{align*}
Ø\neq\{1\}
\end{align*}
Because they are not equal the statement is not true.
The defintion of $f(A)$ for $f:X\to X$ is.
\begin{align*}
f(A)=\{f(x)|x\in A\}\subseteq X
\end{align*}
The definition of $f^{-1}(A)$ for $f:X\to X$ is.
\begin{align*}
f^{-1}(A)=\{x\in X|f(x)\in A\}\subseteq X
\end{align*}
 A: The standard procedure to prove inclusions of sets as $X\subset Y$ is to take an arbitrary element of the included set $x\in X$ and try to show that $x$ has to necessarily belong to Y.
For example, if we are trying to prove a), we will start picking $x\in f(f^{-1}(A))$. What does it mean that $x$ to such a set?
Edit:
Let's tackle c).
When we are trying to show a necessary and sufficient condition, the standard procedure is to first assume the left side and try to prove the right, and then go the other way.
So, let's suppose $f^{-1}(A)=A$. Then to show the left side we have to show that $f(A)\subset A$, since the second conjunct is immediate from our assumption.
As I pointed out earlier, let's pick $x\in f(A)=\{f(z):z\in A\}$. Then by its very definition there is a $z\in A$, such that $f(z) = x$. But as $A=f^{-1}(A)$, $z\in f^{-1}(A)$, so $f(z)=x\in A$ and thus we can conclude that $f(A)\subset A$.
It remains to prove the other implication. Think you can do it?
A: a) $f(f^{-1}(A))\subseteq A$
1) $f(A)=\{f(x)|x\in A\}$
2) $f^{-1}(A)=\{x\in X|f(x)\in A\}$
3) $A = \{A\cap f(X)\} \cup \{A\setminus f(X)\}$
So:
$f(f^{-1}(A)) = \{f(x)|x \in f^{-1}(A)\} = \{f(x)|x \in \{x|f(x) \in A\}\} = \{f(x)|f(x) \in A\} = \{f(x) \in A\} = \{f(x) \in A\cap f(X)\} \cup \{f(x) \in A\setminus f(X)\} =^* \{f(x) \in A\cap f(X)\} = A\cap f(X) \subseteq A$
b) $f(f^{-1}(A))=A$
please note due to equality marked with a star in the previous problem $=^*$ it follows:
$f(f^{-1}(A)) = f(f^{-1}(A\cap f(X))) = A\cap f(X)$
So counterexamples to b) are those and only those in which $A\cap f(X) \neq A$.
That is, b) holds when and only when $A \subseteq f(X)$
c) $f^{-1}(A)=A \Leftrightarrow f(A)\subseteq A$ and $f^{-1}(A)\subseteq A$.
It is necessary because:
$A = f^{-1}(A) \Leftrightarrow f(A) = f(f^{-1}(A)) \subseteq A$
$f^{-1}(A) = A \Leftrightarrow f^{-1}(A) \subseteq A$
It is sufficient, because:
$f^{-1}(f(A)) = \{x \in X|f(x) \in f(A)\} = \{x \in A|f(x) \in f(A)\} \cup \{x \in X\setminus A|f(x) \in f(A)\} = A \cup \{x \in X\setminus A|f(x) \in f(A)\} \supseteq A$
so:
$A \subseteq f^{-1}(f(A)) \subseteq f^{-1}(A)$
That's it.
In the end what is important here is the geometry. 
$f(f^{-1}(A))$ filters out elements not images. No effects when f is surjective, that is when all elements of A are images.
$f^{-1}(f(A))$ puts noise on A, returning the minimal (in the sense of inclusion) cover of A with the preimage subsets. No effects when f is injective, that is when all preimage subsets are singletons, that is when A is made up of unions such singletons.
