The following thought just occured to me on my way back home i forgot it for some while but now i remembered it again is the argument below correct?
Proposition. Lets say that we have a surjective map $f:A\to B$ where $A$ and $B$ are aritrary sets. In addition we have the subsets sets $Y_1,Y_2,...,Y_n$ such that they determine a partition over the set $B$. Does it then follows that the sets $f^{-1}(Y_1),f^{-1}(Y_2),...,f^{-1}(Y_n)$ determine a partion over the set $A$.
I think that the answer is yes and the following is my attempt at a proof is it correct?
Proof. Let $I_n = \{1,2,...,n\}.$ We first prove that $A = \bigcup_{i\in I_n}f^{-1}(Y_i)$. Assume that $a\in A$ consequently $f(a)\in B$ since $B = \bigcup_{i\in I_n}Y_i$ it follows that for some $j\in I_n$ it must be that $f(a)\in Y_j$ equivalently $a\in f^{-1}(Y_j)$ and by implication $a\in\bigcup_{i\in I_n}f^{-1}(Y_i)$. The converse is evident from the fact that $\bigcup_{i\in I_n}f^{-1}(Y_i)$ is in fact a union of subsets of $A$.
We now prove that $f^{-1}(Y_1),f^{-1}(Y_2),...,f^{-1}(Y_n)$ are pairwise disjoint. Assume that for some $i,j\in I_n$ we have $f^{-1}(Y_i)\cap f^{-1}(Y_j)\neq \varnothing$ where $i\neq j$. Since $f^{-1}(Y_i)\cap f^{-1}(Y_j)\neq \varnothing$ we may invoke a $\alpha$ such that $f(\alpha)\in Y_i\cap Y_j$ but $Y_1,Y_2,...,Y_n$ constitute a partition of $B$ and are therefore pairwise disjoint resulting in a contradiction.
Finally since $Y_1,Y_2,...,Y_n$ is partition it follows that $Y_i\neq\varnothing$ which taken together with the surjectivity of $f$ implies that for some $a\in A$ we have $f(a)=b\in Y_i$ equivalently $a\in f^{-1}(Y_i)$ thus $f^{-1}(Y_i)\neq\varnothing$.
$\blacksquare$
Your proof is correct but some details could have been avoided.
Note that $$B= \cup Y_i$$ thus $$A = \bigcup_{i\in I_n}f^{-1}(Y_i)$$
because $x\in f^{-1}(Y_i)$ for some $Y_i$
Also $$ f^{-1}(Y_i)\cap f^{-1}(Y_j) =\phi $$ because $f:A\to B$ is a function.