A description/proof of that preimage by product map is product of preimages by component maps
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 the preimage by any product map is the product of the preimages by the component maps.
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 finite number of sets, \(S_{1, i}\) and \(S_{2, i}\) where \(i = 1, 2, . . ., k\), any corresponding number of maps, \(f_i: S_{1, i} \rightarrow S_{2, i}\), the product map of the maps, \(f_{k + 1}: S_{1, 1} \times S_{1, 2} \times . . . \times S_{1, k} \rightarrow S_{2, 1} \times S_{2, 2} \times . . . \times S_{2, k} = (f_1, f_2, . . ., f_k)\), and any corresponding number of subsets, \(S_{3, i} \subseteq S_{2, i}\), the preimage of the product subsets, \(S_{3, 1} \times S_{3, 2} \times . . . \times S_{3, k}\) by the product map, \(f_{k + 1}^{-1} (S_{3, 1} \times S_{3, 2} \times . . . \times S_{3, k})\), equals the product of the preimages of the subsets by the component maps, \(f_1^{-1} (S_{3, 1}) \times f_2^{-1} (S_{3, 2}) \times . . . \times f_k^{-1} (S_{3, k})\), which is \(f_{k + 1}^{-1} (S_{3, 1} \times S_{3, 2} \times . . . \times S_{3, k}) = f_1^{-1} (S_{3, 1}) \times f_2^{-1} (S_{3, 2}) \times . . . \times f_k^{-1} (S_{3, k})\).
2: Proof
For any element, \(p = (p_1, p_2, . . ., p_k) \in f_{k + 1}^{-1} (S_{3, 1} \times S_{3, 2} \times . . . \times S_{3, k})\), \(f_{k + 1} (p) = (f_1 (p_1), f_2 (p_2), . . ., f_k (p_k)) \in S_{3, 1} \times S_{3, 2} \times . . . \times S_{3, k}\), so, \(f_i (p_i) \in S_{3, i}\), so, \(p_i \in f_i^{-1} (S_{3, i})\), so, \(p = (p_1, p_2, . . ., p_k) \in f_1^{-1} (S_{3, 1}) \times f_2^{-1} (S_{3, 2}) \times . . . \times f_k^{-1} (S_{3, k})\). For any element, \(p = (p_1, p_2, . . ., p_k) \in f_1^{-1} (S_{3, 1}) \times f_2^{-1} (S_{3, 2}) \times . . . \times f_k^{-1} (S_{3, k})\), \(f_{k + 1} (p) = (f_1 (p_1), f_2 (p_2), . . ., f_k (p_k))\), but \(f_i (p_i) \in S_{3, i}\), so, \(f_{k + 1} (p) \in S_{3, 1} \times S_{3, 2} \times . . . \times S_{3, k}\), so, \(p \in f_{k + 1}^{-1} (S_{3, 1} \times S_{3, 2} \times . . . \times S_{3, k})\).
