TITLE: Question from 'How to Prove It'
QUESTION [9 upvotes]: Below is the question from the book mentioned above:

Suppose $f : A \rightarrow B$ and $R$ is an equivalence relation on $A$. We will say that $f$ is compatible with $R$ if $∀x \in A\forall y ∈ A(xRy \rightarrow f(x) = f(y))$.
   Suppose $f$ is compatible with $R$. Prove that there is a unique function
  $h : A/R \rightarrow B$ such that for all $x ∈ A, h\left([x]_R\right)=f(x)$.

Below is my attempt in solving this problem:
Proving the existence of the function:
Let $h=\{(X,y)\in A/R\times B|\exists x\in X(f(x)=y)\}$. Let $X=([x]_R)$ be an arbitrary element of $A/R$ and $x'$ some element of $X$, so $x'Rx$. Since $f$ is compatible with $R$ $(f(x)=f(y))$,so $f(x')=f(x)\in B$. Thus $([x]_R,f(x))\in h$.
Proving the uniqueness of the function:
Let j be a function such that $j([x]_R)=f(x)$. That means we can find some $[x]_R \in A/R$ such that $f(x)=y$. Since $[x]_R\in A/R$ and $y\in B$, $([x]_R,f(x))\in h$, so $j\subseteq h$. By similar reasoning, $h\subseteq j$. So $h$ is unique.
I want to ask that:
1)Is my proof about the existence of the function correct?
2)is my proof about the uniqueness correct? I basically have no idea what I am proving for the uniqueness of the function... 
Please explain and give some hints if there are some mistakes in the proof above. Thanks in advance.

REPLY [2 votes]: There are proofs that make a more or less obvious fact less believable. Yours is of this kind.
There can be at most one function $h:\ A/R\to B$ that satisfies
$$h\bigl([x]_R\bigr)=f(x)\qquad\forall x\in A\ ,\tag{1}$$
since its value on any class $[x]_R$ would be given by $(1)$. 
The real problem is whether $(1)$ actually defines a function on $A/R$. From dealing with equivalence classes we know the following: Given an element $X\in A/R$ there is a representant $x\in A$ of $X$, but this $x$ is not uniquely determined. When both $x$, $x'\in A$ represent $X$ then on the one hand $X=[x]_R=[x']_R$, and on the other hand  $xRx'$. In this case by assumption on $f$ we  have $f(x')=f(x)$, so that $(1)$ defines the same value for $h(X)$, whether we use $x$ or $x'$ to compute it.