Prove that the relation on $X$ given by $x\sim y$ if $f(x)=f(y)$ is an equivalence relation

Let $f:X\to Y$ be a function between two sets. Prove that the relation on $X$ given by $x\sim y$ if $f(x)=f(y)$ is an equivalence relation.

I know that it should have $3$ cases, reflexive (for all $x$ in $X$, $f(x)=f(y)$), symmetric and transitive.

Solution 1:

Reflexivity: Not much to do for this one.

Suppose $f(x) = a,$ then since $f(x) = f(x)$ for all $x \in X$, we have $$f(x) = f(x) \iff a \sim a.$$

Symmetric: Not much to do for this one either.

Suppose $f(x) = f(y)$ and let $f(x) = a$ and $f(y) = b$. Then $$f(x) = f(y) \iff f(y) = f(x).$$ Hence, $$a \sim b \iff b \sim a.$$

Transitive: This is still pretty straightforward, but a bit more involved.

Suppose $f(x) = f(y)$ and $f(y) = f(z)$, and let $a = f(x)$, $b = f(y)$, and $c = f(z)$.

Then $$f(x) = f(y) \implies a = b,$$ and $$f(y) = f(z) \implies b = c.$$

Therefore, $$f(x) = f(y) \textrm{ and } f(y) = f(z) \implies f(x) = f(z) \implies a = c.$$

Hence,

$$ a \sim b \textrm{ and } b \sim c \iff a \sim c$$

Solution 2:

Charles's answer is fine, but to say things a tiny bit differently.

Reflexive: You want to prove that $x\sim x$ for all $x\in X$. This is equivalent to proving that $f(x) = f(x)$ for all $x$ in $X$. But this is obviously true. So the relation is reflexive.

Symmetric: You want to prove that if $x\sim y$, then $y \sim x$. So assume that $x\sim y$. That means that $f(x) = f(y)$. But this is the same as saying that $f(y) = f(x)$ (since equality is symmetric). So we have $y\sim x$. So you have symmetric.

Transitive: You want to prove that if $x\sim y$ and $y\sim z$ then $x\sim z$. So assume the first two. Then you have $f(x) = f(y)$ and $f(y) = f(z)$. But this means that $f(x) = f(y) = f(z)$, or $f(x) = f(z)$. Hence by definition you have $x\sim z$. And you have transitivity.

In all you have proved that the relation is an equivalence relation.