Quotient Mapping is Surjection

Theorem

Let $S$ be a set.

Let $\RR$ be an equivalence relation on $S$.

Then the quotient mapping $q_\RR: S \to S / \RR$ is a surjection.


Proof

Suppose $S$ is empty.

Then, vacuously, $S$ has no equivalence classes

Hence, the set $S / \RR$ of $\RR$-classes of $\RR$ is the empty set.

Therefore, as the codomain of $q_\RR$ is empty, $q_\RR$ is vacuously surjective.

$\Box$


Suppose $S$ is non-empty.

From Equivalence Class is not Empty, we have that:

$\forall \eqclass x \RR \in S / \RR: \exists x \in S: x \in \eqclass x \RR$

and the result follows.

$\blacksquare$


Sources

  • 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts ... (previous) ... (next): Introduction $\S 3$: Equivalence relations
  • 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 6$. Indexed families; partitions; equivalence relations
  • 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 7$: Relations
  • 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.4$: Equivalence relations
  • 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): Appendix $\text{A}$ Preliminaries: $\S 1.$ Linear Algebra
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