Can an equivalence relation $C$ on $A$ relate two non-equal elements of a set $A$?

Question

I was working through one of the exercises in Topology: A First Course by Munkres, and I came across this:

Let $f : A \to B$ be a surjective function. Let us define a relation on $A$ by setting $a_0 C a_1$ if $f(a_0) = f(a_1)$. Show that $C$ is an equivalence relation.

Now essentially what this is saying is the following:

$$C = \{(a, b) : f(a) = f(b) \ \ \text{where} \ \ a,b \in A\}$$

But since $f$ is surjective and not injective we could have $f(a) = f(b)$ when $a \neq b$. But clearly by what we have above $aCb$, where $C$ is an equivalence relation on $A$.

Is this an error or a misunderstanding on my part or can an equivalence relation relate two elements of a set that are not in fact equal to each other?

Answer 1

The point of equivalence relations is to generalize the idea of equality. There's only one equivalence relation on a set that is the equality relation, and every other one is not and necessarily declares two unequal elements to be "equivalent".

I would recommend cracking open a discrete math/intro to proofs book and reading about how equivalence relations are connected to partitions on a set.

Answer 2

Yes, an equivalent relation can, and usually does, relate elements that are not originally equal to each other, generalizing what we know as equality.

Let us take, for instance, $\mathbb{Z}$ with the following equivalence relation: $x \sim y$ iff $x - y$ is even. With the above relation, we generalize the idea of "equality" saying that two numbers are "equal" if they are both even or both odd, e. g., $1 \equiv 3$.