392: Composition of Preimage After Map of Subset Is Identical Iff It Is Contained in Argument Set

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A description/proof of that composition of preimage after map of subset is identical iff it is contained in argument 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
• 3: Note

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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, the composition of the preimage after the map of any subset is identical if and only if it is contained in the argument set.

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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, $f:S_1\to S_2$, and any subset, $S_3\subseteq S_1$, $f^{-1}\circ f(S_3)=S_3$ if and only if $f^{-1}\circ f(S_3)\subseteq S_3$.

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2: Proof

Suppose that $f^{-1}\circ f(S_3)\subseteq S_3$. For any $p\in f^{-1}\circ f(S_3)$, $p\in S_3$. For any $p\in S_3$, $f(p)\in f(S_3)$, which means that $p\in f^{-1}\circ f(S_3)$ by the definition of preimage.

Suppose that $f^{-1}\circ f(S_3)=S_3$. $f^{-1}\circ f(S_3)\subseteq S_3$.

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3: Note

It is important not to carelessly conclude that $f^{-1}\circ f(S_3)=S_3$ without checking the condition.

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References

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