Do total orderings of equivalence classes of subsets also totally order the same equivalence classes of the set.

Question

$A\cup B=C$

Suppose some function $f$ totally orders equivalence classes of $A$ and equivalence classes of $B$, both of which are formed by the same equivalence relation $\sim$.

The domain and range of $f$ is $C$

The range of $f$ in $A$ is $A$

The range of $f$ in $B$ is $B$

Does this imply that $f$ totally orders equivalence classes of $C$ formed by the same equivalence relation?

Answer 1

For a counter-example let $C=\{1,2,3,4\} , A=\{1,2\} $ and $B=\{3,4\}.$ Let the set of equivalence classes be $E= \{\{1,3\} , \{2,4\}\}.$

By "the equivalence classes of $A$" I assume you mean $\{x\cap A: x\in E\}$ \ $\{\phi\}.$ So the equivalence classes of $A$ are $\{1\}$ and $\{2\}.$ And the equivalence classes of $B$ are $\{3\}$ and $\{4\}.$

Suppose $\{1\}<\{2\}<\{4\}<\{3\}.$ This doesn't "translate" to a linear order on $E.$

Answer 2

Not necessarily. We could have $f$ map an equivalence class from $A$ and a different equivalence class from $B$ to the same element in your order and that would break the totality.

If $f$ is one-to-one or $f|_A$ and $f|_B$ have disjoint ranges, that would give a total order.