Let $f: S \to T$ be a surjective function that maps the elements of each subset $S_{\alpha} \subset S$ to elements $t \in T$. The problem is to show that the collection $\{f^{-1}(\{t\}) = S_{\alpha}\}$ for each $t \in T$ partition the original set $S$.
I can show that the $\bigcup S_{\alpha} = S$.
How do I show that if $S_{\alpha} \neq S_{\beta}$, then $S_{\alpha} \cap S_{\beta} = \emptyset$?
Write $S_a = f^{-1}(t_a)$ and $S_b = f^{-1}(t_b)$. So $S_a \neq S_b$ means $t_a \neq t_b$. Now if $s \in S_a \cap S_b$ then $f(s) = t_a \neq t_b = f(s)$. So the element $s$ is mapped to two different elements, which is not allowed for a function.