A description/proof of that maps composition preimage is composition of map preimages in reverse order
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 maps composition, the preimage under the composition is the composition of the map preimages in the reverse order.
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, \(S_1, S_2, . . ., S_n\) where \(3 \le n\), any maps, \(f_1: S_1 \rightarrow S_2, f_2: S_2 \rightarrow S_3, . . ., f_{n - 1}: S_{n - 1} \rightarrow S_n\), and any subset, \(S \subseteq S_n\), \((f_{n - 1} \circ . . . \circ f_2 \circ f_1)^{-1} (S) = {f_1}^{-1} ({f_2}^{-1} (. . . ({f_{n - 1}}^{-1} (S)) . . .))\).
2: Proof
Suppose that \(n = 3\). For any element, \(p \in (f_2 \circ f_1)^{-1} (S)\), \(f_2 \circ f_1 (p) \in S\), \(f_1 (p) \in {f_2}^{-1} (S)\), \(p \in {f_1}^{-1} ({f_2}^{-1} (S))\). For any \(p \in {f_1}^{-1} ({f_2}^{-1} (S))\), \(f_1 (p) \in {f_2}^{-1} (S)\), \(f_2 \circ f_1 (p) \in S\), \(p \in (f_2 \circ f_1)^{-1} (S)\).
Suppose that \((f_{n - 1} \circ . . . \circ f_2 \circ f_1)^{-1} (S) = {f_1}^{-1} ({f_2}^{-1} (. . . ({f_{n - 1}}^{-1} (S)) . . .))\) for an \(n\). \((f_{n} \circ f_{n - 1} \circ . . . \circ f_2 \circ f_1)^{-1} (S) = (f_{n} \circ (f_{n - 1} \circ . . . \circ f_2 \circ f_1))^{-1} (S) = {(f_{n - 1} \circ . . . \circ f_2 \circ f_1)}^{-1} ({f_{n}}^{-1} (S)) = {f_1}^{-1} ({f_2}^{-1} (. . . ({f_{n - 1}}^{-1} ({f_{n}}^{-1} (S))) . . .))\).
So, by mathematical induction, the proposition holds for any \(3 \le n\).
