This is a problem from Proofs and Fundamentals by Ethan D. Bloch that I’m struggling to solve:
Let $f:A \to B$ a map. Define a relation $\sim$ on $A$ by letting $x \sim y$ iff and only if $f(x) = f(y)$, for all $x, y \in A$. What can be said about the equivalence classes of $\sim$, depending upon whether $f$ is injective but not surjective, surjective but not injective, neither or both?
So far, I realize that if $f$ is injective, then all the equivalence classes of $\sim$ will have exactly one element (and this comes also from the fact that $f$ is a map).
Although, I don’t think that the fact that $f$ is surjective or not will alter the equivalence classes. Can someone please help me understanding what is the “relation” (if any) between surjectivity and equivalence classes?
Thank you in advance for your attention!
Best Answer
Since $f$ is surjective we see that for each $b\in B$ preimage $f^{-1}(b)$ is nonempty. Equivalence classes are like : $$[a] = \{x\in f^{-1}(b); b=f(a)\}$$
Since it is surjective we can only say that quotien set $A/_\sim$ and $B$ are set isomorphic (i.e. there is bijection between them).