Is contradictory that in a mapping two disjoints sets have the same preimage? Suppose $f:A \longrightarrow B $ and $B=X \cup Y$, with $X\cap Y=\emptyset $. Also, $f^{-1}(X)=f^{-1}(Y)=A$. Is this contradictory? I think if that is the case, then there would exist $y_1\in X$, $ y_2\in Y$ such that $f(x)=y_1,y_2$ for some $x\in A$, but this would be impossible because $f$ is a mapping.

The situation you describe is not contradictory: you can take $A$, $B$, $f$, $X$ and $Y$ all empty. If any of these sets is non-empty, then you get a contradiction. Your argument is assuming that one of the sets is non-empty, but there is nothing in the statement of the problem to justify that assumption.