Proving that if $E, F$ are equivalence relations on $A$ and $E \subseteq F$, then there is a surjection from $A\setminus E$ to $A\setminus F$

Question

Proving that if $E, F$ are equivalence relations on $A$ and $E \subseteq F$, then there is a surjective function from $A\setminus E$ onto $A\setminus F$.

What does $E \subseteq F$ even mean? Does it mean that $xEy \rightarrow xFy$? Can't seem to have a direction on how to solve this.

Thanks in advance!

Answer 1

If $E$ and $F$ are equivalence relations on $A$, then $E \subseteq F$ mean that for for all $x,y \in A$, $x \ E \ y$ implies $x \ F \ y$. Thus for all $a \in A$, $[a]_E \subseteq [a]_F$. Also observe that each $F$-class is a union of $E$-classes.

Also equivalence relations are partitions of the space $A$. $E \subseteq F$ means that $E$ is a finer partition of $A$.

So suppose $C \in A \backslash E$. Then $C = [x]_E$ for some $x \in A$. Define $\Phi(C) = [x]_F$. You can check that $\Phi$ is well-defined and surjective from $A \backslash E \rightarrow A \backslash F$.

Answer 2

As equivalence relations in $A$, the sets $E,F$ are subsets of $A\times A$, hence speaking of $E\subseteq F$ makes sense. As $(x,y)\in E$ then implies $(x,y)\in F$, this translates to $xEy\to xFy$.

The only way to get a map $A/E\to A/F$ in any obvious manner is to map the $E$-equivalence class of $(x,y)$ to the $F$-equivalence class of $(x,y)$. The condition $E\subseteq F$ implies that this map is well-defined (in fact is logically equivalent to it). It is clearly onto.