Questions about fibers (pre-image) Just want to check if I have done this correctly. Elegant, simple, concise answers are welcome and preferred. It took me a long time to get this far.
Problem statement: Let $f: A → B$ be a function. Let $\overline{A}$ be the set of non-empty fibres of $f$. Let $l(a)$ be the unique fibre of $f$ which contains $a$; this defines a function $l: A → \overline{A}$. 
(a) Is $l$ always going to be surjective, or does it depend on the choice of the $f$ that we started with? What about injective?
Define $\overline{f}:\overline{A}→B$ by $\overline{f}(l(a))=f(a)$.
(b) Show that $\overline{f}$ is injective.
(c) Show that $\overline{f}$ uniquely occupies its position in the equation $f=\overline{f}∘l$. In other words, let $F: \overline{A}→B$ be any function such that $f=F∘l$, and show from this that $F=\overline{f}$.
My solution: 
(a) Yes (to surjectivity), if $a$ is in $A$, then $f(a)$ is in $B$. So every $a$ is a preimage of some $b$. No (to injectivity). Consider $A$ consisting of the elements 1 and 2, $B$ consisting of the element 3, with $f(1)=3,f(2)=3$. Then $l(1)=$ the set consisting of the elements 1 and 2 and $l(2)=$ the set consisting of the elements 1 and 2.
(b) $\overline{f}(l(a))= \overline{f}(l(b))$ which implies that $f(a)=f(b)$. This implies that $a$ and $b$ are in the same fibre of $f$, say the fibre of $f$ at $c$. So then, $l(a)=l(b)$.
(c) $f=\overline{f}∘l=F∘l$. Since $\overline{f}$ is injective, $f^{-1}$ exists. So, $\overline{f}^{-1}∘(F∘l)=\overline{f}^{-1}∘(\overline{f}∘l)$. So $(\overline{f}^{-1}∘F)∘l=l$, which implies that $\overline{f}^{-1}∘F =$ identity function. $\overline{f}∘\overline{f}^{-1}∘F=\overline{f}(x)$. Thus $F(x)=\overline{f}(x)$, and thus $F=\overline{f}$.
 A: Yes, it’s right; properly speaking one ought to show that $\bar f$ is well-defined, but you weren’t asked to do this. 
The last part can be done more easily. Suppose that $F\circ\ell=f=\bar f\circ\ell$, and let $\alpha\in\overline A$ be arbitrary. By definition $\alpha=\ell(a)$ for some $a\in A$, so 
$$F(\alpha)=F\big(\ell(a)\big)=(F\circ\ell)(a)=f(a)=(\bar f\circ\ell)(a)=\bar f\big(\ell(a)\big)=\bar f(\alpha)\;;$$
$\alpha\in\overline A$ was arbitrary, so $F=\bar f$.

You might want to note that this is yet another way of thinking about the cluster of ideas surrounding equivalence relations, partitions, and quotients by equivalence relations. For $x,y\in A$ write $x\sim y$ iff $f(x)=f(y)$, and observe that this is an equivalence relations. The equivalence classes of $\sim$ are the fibres of $f$, which therefore form a partition $\overline A$ of $A$. Moreover, $\overline A=A/\sim$, the quotient of $A$ by the relation $\sim$, and $\ell:A\to\overline A$ is the quotient map. The quotient map $\ell$ collapses each fiber (member of the partition, equivalence class) to a single point of $\overline A$, and $\bar f$ then acts on those points just as $f$ acted on the points of $A$ that were collapsed to form them.
