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

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Topics

About: set

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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 map.

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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.

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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, $S_1$ and $S_2$, any map, $f:S_1\to S_2$, 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 $f^{-1}(S_2\setminus S_3)=S_1\setminus f^{-1}{S}_3$.

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

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References

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