Reasoning behind intuitive understanding of specific equivalence class $a \sim b \iff f(a) = f(b)$

Question

I was attempting this question the other day:

Let $f:A \rightarrow B$ be a surjective map of sets. Prove the relation $$a \sim b \iff f(a) = f(b)$$ is an equivalence relation whose equivalence classes are the fibres of $f$.

In principle this is easy to solve. Reflexivity, Symmetry and Transitivity are all straightforward since

• $f(a) = f(a)$
• $f(a) = f(b) \iff f(b) = f(a)$
• $f(a) = f(b) \wedge f(b) = f(c) \rightarrow f(a) = f(c)$

Any fibre of $f$ is $\{a\in A|f(a) = b\}$ and since the right inverse $h$ exists, $b$ can be written as $f(c)$. $f(a) = f(c) \iff a\sim b$. Simple enough? Except this all sounded a little too straightforward and I couldn't really understand in particular how the equivalence classes are the same as the fibres of $f$. So I thought up this lil visual representation.

All the elements of $B$ are occupied i.e. the codomain is the range. Sometimes multiple elements of $A$ point towards a single element of $B$. That is allowed by our required surjectivity. Now every time $f(x) = f(y)$ for two elements $x,y \in A$, $x \sim y$. $x$ and $y$ share an equivalence class by this definition. I feel this captures the intrinsic symmetry of equivalence classes well. $x$ and $y$ are "equivalent" in that there is a certain indistinguishability about them. They are identical in the one way we care about, that applying $f$ to either yields the same result. Within that little coloured bubble inside $A$, they are all the same. I think this best shows how the equivalence classes are the fibres of $f$.

P.S. you can also see how all the equivalence classes form a partition of $A$.

Am I correct in everything above? In particular, I still do not understand why $f$ must be surjective because while it was an integral part of the proof, I could always add a few extra harmless elements to $B$, invalidating the surjective property of $f$, and the equivalence classes would remain the fibres of $f$.

Answer 1

If $f$ is not surjective, then the fibres of $B$ elements that are not the images of $A$ elements under $f$ would have empty fibres, whereas equivalence classes are non-empty by definition. The empty fibres would therefore not be equivalence classes of $f$ – "whose equivalence classes are the fibres of $f$" implies that the result also requires fibres to be equivalence classes.