2022-08-07

332: Preimage Under Domain-Restricted Map Is Intersection of Preimage Under Original Map and Restricted Domain

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A description/proof of that preimage under domain-restricted map is intersection of preimage under original map and restricted domain

Topics


About: set
About: map

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that for any map between sets and its any domain-restriction, the preimage under the domain-restricted map is the intersection of the preimage under the original map and the restricted domain.

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, \(S_1, S_2\), any map, \(f: S_1 \rightarrow S_2\), any subsets, \(S_3 \in S_1, S_4 \in S_2\), and the domain-restriction, \(f|_{S_3}: S_3 \rightarrow S_2\), \({f|_{S_3}}^{-1} (S_4) = f^{-1} (S_4) \cap S_3\).


2: Proof


For any \(p \in {f|_{S_3}}^{-1} (S_4)\), \(f|_{S_3} (p) = f (p) \in S_4\) by the proposition that for any map between sets, the image of any point is on any subset if and only if the point is on the preimage of the subset, so, \(p \in f^{-1} (S_4)\). Of course, \(p \in S_3\), so, \(p \in f^{-1} (S_4) \cap S_3\).

For any \(p \in f^{-1} (S_4) \cap S_3\), \(f (p) \in S_4\) by the proposition that for any map between sets, the image of any point is on any subset if and only if the point is on the preimage of the subset. As \(p \in S_3\), \(f (p) = f|_{S_3} (p) \in S_4\), so, \(p \in {f|_{S_3}}^{-1} (S_4)\).


References


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