A description/proof of that for bijection, preimage of subset under inverse of map is image of subset under map
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of set.
- The reader knows a definition of map.
- The reader admits the proposition that for any map, the composition of the preimage after the map of any subset is identical if the map is injective with respect to the argument set image.
Target Context
- The reader will have a description and a proof of the proposition that for any bijection, the preimage of any subset under the inverse of the map is the image of the subset under the map.
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.
Main Body
1: Description
For any sets, \(S_1\) and \(S_2\), any bijection, \(f: S_1 \rightarrow S_2\), and any subset, \(S_3 \subseteq S_1\), the preimage of \(S_3\) under the inverse of \(f\) is the image of \(S_3\) under \(f\), which is \((f^{-1})^{-1} (S_3) = f (S_3)\).
2: Proof
For any \(p \in (f^{-1})^{-1} (S_3)\), \(f^{-1} (p) \in S_3\), \(p \in f (S_3)\); for any \(p \in f (S_3)\), \(f^{-1} (p) \in f^{-1} (f (S_3)) = S_3\) where the left \(f^{-1}\) is the operation of the inverse while the right \(f^{-1}\) is the preimage, by the proposition that for any map, the composition of the preimage after the map of any subset is identical if the map is injective with respect to the argument set image, \(p \in (f^{-1})^{-1} (S_3)\).
