A description/proof of that map preimage of codomain minus set is domain minus preimage of set
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of map.
Target Context
- The reader will have a description and a proof of the proposition that the preimage of the codomain minus any codomain subset under any map is the domain minus the preimage of the subset.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any sets, \(S_1\) and \(S_2\), any map, \(f: S_1 \rightarrow S_2\), and any subset of the codomain, \(S_3 \subseteq S_2\), the preimage of the codomain minus the subset is the domain minus the preimage of the subset, which is \(f^{-1} (S_2 \setminus S_3) = S_1 \setminus f^{-1} {S_3}\).
2: Proof
Suppose \(p \in f^{-1} (S_2 \setminus S_3)\). \(f (p) \in S_2 \setminus S_3\), which means that \(p \in S_1\) but \(p \notin f^{-1} (S_3)\), which is \(p \in S_1 \setminus f^{-1} (S_3)\). Suppose \(p \in S_1 \setminus f^{-1} (S_3)\). \(f (p) \in S_2\) but \(f (p) \notin S_3\), which means \(f (p) \in S_2 \setminus S_3\), which is \(p \in f^{-1} (S_2 \setminus S_3)\).
