Given $f: A \to B$ a surjective function, prove that $f$ determines a partition on $A$ ($A$ and $B$ are abstract sets where $|B| = n$)

Question

Given $f: A \to B$ a surjective function, prove that $f$ determines a partition on $A$ ($A$ and $B$ are abstract sets where $|B| = n$).

I know to show the inverse $f(b)$ leads to $A$, etc. What I don't know how to do is show that the union of all the partitions is the set A.

Answer 1

If $B = \{b_1,\ldots,b_n\}$ with $b_j\neq b_k$ for $j\neq k$, then $A = \bigcup_{k=1}^nf^{-1}(\{b_k\})$, which is a union of mutually disjoint sets and thus a partition of $A$.

Disjointness: If $a\in f^{-1}(\{b_j\})\cap f^{-1}(\{b_k\})$ for $j\neq k$, then $f(a) = b_j$ and $f(a) = b_k$, which is not possible.

Equality: First, it is clear that $f^{-1}(\{b_k\})\subset A$ for each $k$, so the union is a subset of $A$. Conversely, if $a\in A$, then $f(a) = b_k$ for some $k$, hence $a\in f^{-1}(\{b_k\})$ and so $a$ is contained in the union.

BTW: $f$ does not have to be surjective for this to be true.