Composition of function with it's inverse on subdomains I have a short question. We have to check the following statements and tell for which one the equal sign holds.

Let $M \subset \mbox{domain } f$ and $N \subset \mbox{Im } f$.
$$\{f(f^{-1}(t)\mid t \in N\} \subset N$$ $$ M \subset \{f^{-1}(f(t))\mid t \in M\}$$

I don't get it as $f(f^{-1}(t)) = t$ so $\{t\mid t \in N\} = N$. The same is true for $f^{-1}(f(t)) = t$ so $\{t\mid t \in M\} = M$.
For me it seems to be trivial and both statements are true as well as equal to the subsets. What did I do wrong? Can anyone help?
 A: Let's have a look at what the inverse of $f$ actually means. The definition of $f^{-1}(t)$ is the set $^{[1]}$ of $s$ such that $f(s)=t$. The important point to notice here is that this doesn't need to be a set with a single element; it can have zero elements, one element, or multiple elements. For instance, if $f$ is not a surjective function, then there exists some $t\in\mbox{codomain }f$ such that there is no $s$ with $f(s)=t$, and so $f^{-1}(t)=\emptyset$.
Let's apply this to your second statement then to see why this does not give set equality as you suspect. Let $f\colon \{0,1\}\rightarrow \{a\}$ be given by $f(0)=f(1)=a$. We'll let $M=\{0\}$ and $N=\mbox{Im } f$ for the purposes of this example. Now we see that $$\begin{array}{rcl}\{f^{-1}(f(t))\mid t\in M\}&=&\{f^{-1}(f(0))\}\\
&=&\{f^{-1}(a)\}\\
&=&\{0,1\}\neq\{0\}\end{array}$$
and so equality doesn't hold for the second statement. This happened because $f$ is not injective.
For the first statement, it turns out that you will get equality, but only because $N\subset \mbox{Im }f$. If $N$ was instead a subset of the codomain of $f$ (in general a bigger set than the image), and $f$ was not surjective, then the first statement would not hold as equality either (take $f\colon\{0\}\rightarrow\{a,b\}$ with $f(0)=a$ and let $M=\{0\}$, $N=\{a,b\}$).
Now that you realise the problem isn't totally trivial, and really does require a proof, I'll leave that part to you (as it seems your question was more asking why equality doesn't always hold). If you need help with that proof, feel free to leave a comment and I can elaborate.

Following the OP's comment, I'll give a proof of the first statement, hopefully you can then prove the second statement yourself.
Let $f\colon X \rightarrow Y$ be a function of sets and let $M\subset\mbox{domain }f\subset X$ and $N\subset\mbox{Im }f\subset Y$.
We wish to show that $A=\{f(f^{-1}(t))\mid t\in N\}\subset N$. Let $s\in A$ and so by definition there exists a $t\in N$ such that $f(f^{-1}(t))=s$. Note that $f^{-1}(t)=\{u\in X\mid f(u)=t\}$, and so in particular, for every $u\in f^{-1}(t)$, we have $f(u)=t$. it follows that $f(f^{-1}(t))=t$ and so $s=t$. Hence because $t$ is in $N$, we have $s\in N$. As $s$ was chosen arbitrarily, this implies that $A\subset N$ as required.

[1] Even though the inverse of $f$ is a set, it seems that the statements in your question don't take this in to account as otherwise the sets in question would not contain elements in common with $N$ and $M$ respectively because they would be subsets of the powersets of $\mbox{codomain }f,\mbox{domain }f$ respectively. Given that the problem statement then is being rather liberal with notation, I'll also use the abuse in the rest of the answer.
