A description/proof of that composition of preimage after map of subset is identical if map is injective with respect to argument set 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 preimage after the map of any subset is identical if the map is injective with respect to the argument set 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, \(S_3 \subseteq S_1\), \(f^{-1} \circ f (S_3) = S_3\) if \(f\) is injective with respect to \(f (S_3)\), which means that for any \(p \in f (S_3)\), \(f^{-1} (p)\) is a 1 element set.
2: Proof
Suppose that \(f\) is injective with respect to \(f (S_3)\). For any \(p \in f^{-1} \circ f (S_3)\), \(f (p) \in f (S_3)\) by the definition of preimage, which means that there is a \(p' \in S_3\) such that \(f (p') = f (p)\), but as \(f^{-1} (f (p))\) is a 1 element set, \(p' = p\), so, \(p \in S_3\). 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.
3: Note
It is important not to carelessly conclude that \(f^{-1} \circ f (S_3) = S_3\) without checking the condition.
The condition is not any necessary condition; in fact, the issue is not really being injective, but \(f^{-1} \circ f (S_3) \subseteq S_3\), which is a necessary condition; being injective is just a typical case of guaranteeing \(f^{-1} \circ f (S_3) \subseteq S_3\).
The condition is 'with respect to \(f (S_3)\)', not "with respect to \(S_3\)": the condition is not about the injectivity of \(f|_{S_3}: S_3 \rightarrow S_2\).
