2022-05-01

66: Map Preimage of Codomain Minus Set Is Domain Minus Preimage of Set

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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



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 \(p \in f^{-1} (S_2 \setminus S_3)\).


References


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