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

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Topics

About: set
About: map

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of set.
• The reader knows a definition of map.

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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.

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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.

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Main Body

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1: Description

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

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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, $p\in {f{|}_{S_3}}^{-1}(S_4)$.

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References

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