A description/proof of that composition of map after preimage is identical iff argument set is subset of map image
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.
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 image.
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, \(S_1, S_2\), any map, \(f: S_1 \rightarrow S_2\), and any subset, \(S_3 \subseteq S_2\), \(f \circ f^{-1} (S_3) = S_3\) if and only if \(S_3 \subseteq f (S_1)\).
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 \(S_3 \subseteq f (S_1)\), \(f^{-1} (p) \neq \emptyset\), 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 \(p \notin f (S_1)\). \(f^{-1} (p) = \emptyset\), which would mean that \(p \notin f \circ f^{-1} (S_3)\), so, \(p \notin S_3\), a contradiction.
3: Note
It is important not to carelessly conclude that \(f \circ f^{-1} (S_3) = S_3\) without checking the condition.
