Prove that $\{A,B\}$ and $\{C,D\}$ partition for $X$ and $Y$ (in order).

Question

Question: Suppose $f:X\to Y$ and $g:Y\to X$ two arbitrary functions. Prove that $\{A,B\}$ is a partition for X and $\{C,D\}$ is a partition for Y such that :
$$\begin{align} f(A) = C\end{align}$$ $$\begin{align} g(D) = B\end{align}$$ Definitions that I know:
I have to show that $\bigcup_{i\in I} \{A,B\}_{i\in I} = X$ and $\bigcup_{j\in J} \{C,D\}_{j\in J} = Y$ and also if $i\neq j$ then $\{A,B\}_{i\in I}\cap \{A,B\}_{j\in J} = \varnothing$.
I know that by Axiom schema of specification and Axiom of pairing there is set $$\begin{align} D=\{A,B\} = \{x\in C : x=A \,\text{ or }\, x=B\}\end{align}$$ I have know idea how to start, just clear for what's going on? I want to show that there are exists such that partitions.

Answer 1

First note that by definition a partition of $X$ is any choice of two nonempty sets $A,B$ which are disjoint, and whose union is $X.$ Simalarly for the partition of $Y$ into two sets $C,D.$ The question is whether, for any sets $X,Y$ and any functions $f:X \to Y,\ g:Y \to X$ there are partitions of $X$ into $A,B$ and of $Y$ into $C,D$ such that $$f(A)=C,\ \ f(D)=B.$$

The point of this answer is to show the answer to the question posed is no. For this we let $X$ denote the set of nonzero integers, and $Y$ the set of positive integers. We define $f:X \to Y$ by $f(x)=|x|,$ the absolute value of the integer $x.$ And we define $g:Y \to X$ by $g(y)=y,$ which we may do since any positive integer is also a nonzero integer and so is in set $X.$ Now $A$ is a collection of nonzero integers. First we note that it cannot be the case that, for each nonzero integer $x,$ our set $A$ contains either $x$ or $-x$ (or both). This is because if it did then every $y$ in $Y$ would be $g(x),$ i.e. $C=g(X)$ would be the entire set $Y.$ Tht is not possible because the complement of $C$ in $Y$ is by definition $D,$ so $D$ would be empty, while we need $D$ nonempty in order for $C,D$ to be a partition of $Y.$

So we know there is a specific pair $-e,e$ of integers neither of which lies in $A.$ It follows we have $-e,e \in B$ since $A,B$ partitions $X.$ Also the positive integer $k=|e|$ is not in $C,$ so it must lie in $D.$ But then we do not have $g(D)=B$ because $B$ has in it the negative integer $-k$ which is not $g(y)$ for any $y$ in $D.$ Thus we have arrived at a contradiction and the desired sets $A,B,C,D$ do not exists in our example.