functions - Problem 4 from Section E of Chapter 12 of Pinter's Book of Abstract Algebra

The question: Let $f: A\rightarrow B$ be a function, and let $\{ B_i : i \in I\}$ be a partition of $B$. Prove that $\{f^{-1}(B_i):i \in I\}$ is a partition of $A$.

My work so far: For any given $B_i \in \{ B_i : i \in I\}$, the symbol $f^{-1}(B_i)$ denotes the preimage of $B_i$. The preimage of $B_i$ is the set $\{x \in A: f(x) \in B_i\}$. Since $\{B_i : i \in I\}$ is a partition of $B$, each $b \in B$ belongs to exactly one $B_i$. Also, because $f:A \rightarrow B$ is a function, if $x \in A$ then $f(x) \in B$. Thus if $x \in A$ then $f(x) \in B_i$ for exactly one $B_i$.

What I need help with: In order to complete the proof, I still need to show that each $f^{-1}(B_i)$ is nonempty. This seems impossible to me, because if you have a $B_i$ whose elements are all outside the image set of $A$ then the preimage of $B_i$ would be the empty set. The problem does not say that $f$ must be surjective, so I don't think that can be assumed either. So, this leads me to believe that I am wrong in assuming that $f^{-1}(B_i)$ refers to the preimage of $B_i$. Am I indeed wrong? If so, what does $f^{-1}(B_i)$ mean? If not, how can I prove that all $f^{-1}(B_i)$ are nonempty?

1 Answer

  1. Brent- Reply


    I think Pinter got sloppy here. You do get pairwise disjoint subsets of $A$ whose union is all of $A$, but not all of those are non-empty.

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