Any function induces a surjection by restricting its codomain to its range. This should be obvious. What may or may not be obvious is that any onto function induces a bijection defined on a quotient of its domain. Let f: A → B be an onto function. Let Q be the set of all equivalence classes of domain A under the equivalence relation x ~ y if and only if f(x) = f(y). (Equivalently, Q is the set of all pre-images under f).
Prove that ~ is an equivalence relation.

Question

Any function induces a surjection by restricting its codomain to its range. This should be obvious. What may or may not be obvious is that any onto function induces a bijection defined on a quotient of its domain. Let f: A → B be an onto function. Let Q be the set of all equivalence classes of domain A under the equivalence relation x ~ y if and only if f(x) = f(y). (Equivalently, Q is the set of all pre-images under f).
 Prove that ~ is an equivalence relation.

swissgirl · Answer

nnbboouuusssccaaallllllll Long time no see

anonymous · Answer

For this proof we will need to show reflexive, symmetric, and transitive. They're all pretty easy, because it's just equality under the function. Reflexive: f(x)=f(x), Symmetric: f(x)=f(y) implies f(y)=f(x), Transitive: f(x)=f(y) and f(y)=f(z) implies f(x)=f(z).

swissgirl · Answer

what does ~ mean?

swissgirl · Answer

relation?

anonymous · Answer

Tilde is the standard notation for an equivalence relation. You can use it for any relation, but usually it's used specifically for equivalence relations.

swissgirl · Answer

thhaank youuuuuuu