How to prove that a1 ∼ a2 ⇔ f(a1) = f(a2) is an equivalence relation?

Question

Suppose a function f : A → B is given. Define a relation ∼ on A as follows: a1 ∼ a2 ⇔ f(a1) = f(a2).

a) Prove that ∼ is an equivalence relation on A.

I know that I have to prove for the reflexive, symmetric, and transitive properties, but how do I do that?

Answer 1

Denote the relation given by $aRb$

Reflexivity

Clearly $f(a_1)=f(a_1)$, then $a_1R$ $a_1$

Simetry

If $a_1R$ $a_2$, then $f(a_1)=f(a_2)$, consequently $f(a_2)=f(a_1)$ because $=$ is a symmetric relation, i.e, $a_2R$ $a_1$.

Transitivity

If $a_1R$ $a_2$ and $a_2R$ $a_3$, then $f(a_1)=f(a_2)$ and $f(a_2)=f(a_3)$, consequently $f(a_1)=f(a_3)$, and then $a_1R$ $a_3$.

For the properties from above $R$ is an equivalence relation.

Answer 2

(i) Reflexivity: If $a$ ~ $a$, then clearly $f(a)=f(a)$. Likewise, if $f(a)=f(a)$, then $a=a$.

(ii) Symmetry: If $a$ ~ $b$ then $f(a)=f(b)$. Likewise we have $b$ ~ $a$, since for any thing, $c=d$ implies $d=c$.

(iii) Transitivity: If $a$ ~ $b$ then $f(a)=f(b)$. And if $b$ ~ $c$ then $f(b)=f(c)$. And by the properties of $=$ we have $f(a)=f(c)$, that is $a$ ~ $c$. And conversely, we see that $f(a)=f(b)=f(c)$ implies $a$ ~ $b$, $b$ ~ $c$ and $a$ ~ $c$.

Here is link that will be useful: http://en.wikipedia.org/wiki/Equivalence_relation