329: Point Is on Map Image of Subset if Preimage of Point Is Contained in Subset, but Not Only if

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A description/proof of that point is on map image of subset if preimage of point is contained in subset, but not only if

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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
• 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 between sets, any point is on the image of any subset if the preimage of the point is contained in the subset, but not only if.

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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 point, $p\in S_2$, and any subset, $S_3\subseteq S_1$, $p\in f(S_3)$ if $f^{-1}(p)\subseteq S_3$, but not only if.

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2: Proof

Suppose that $f^{-1}(p)\subseteq S_3$. $f(f^{-1}(p))=\{p\}\subseteq f(S_3)$, so, $p\in f(S_3)$.

Suppose that $p\in f(S_3)$. If $f$ is not injective, there may be multiple points on $f^{-1}(p)$, one of which may not be on $S_3$.

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3: Note

We have to be careful not to assume the reverse, by confusing with another 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.

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References

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