169: Composition of Map After Preimage Is Identical iff Argument Set Is Subset of Map Range

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description/proof of that composition of map after preimage is identical iff argument set is subset of map range

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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 map after any preimage is identical if and only if the argument set is a subset of the map range.

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

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2: Proof

Suppose that $S_3\subseteq f(S_1)$. For any $p\in f\circ f^{-1}(S_3)$, $p\in S_3$ by the definition of preimage. For any $p\in S_3$, as $S_3\subseteq f(S_1)$, $f^{-1}(p)\ne \mathrm{\varnothing }$, and $f(f^{-1}(p))=p$ by the definition of preimage, so, $p\in f\circ f^{-1}(S_3)$.

Suppose that $f\circ f^{-1}(S_3)=S_3$. Suppose that it was not that $S_3\subseteq f(S_1)$. There would be a $p\in S_3$ such that $p\notin f(S_1)$. $f^{-1}(p)=\mathrm{\varnothing }$, which would mean that $p\notin f\circ f^{-1}(S_3)$, so, $p\notin S_3$, a contradiction.

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3: Note

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

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References

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