A description/proof of that some para-product maps of continuous maps are continuous
Topics
About: topological space
About: map
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Description 1
- 2: Proof 1
- 3: Note 1
- 4: Description 2
- 5: Proof 2
Starting Context
- The reader knows a definition of product topology.
- The reader knows a definition of continuous map.
- The reader admits a characterization of product topology.
- The reader admits the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
- The reader admits the proposition that the preimage by any product map is the product of the preimages by the component maps.
- The reader admits the proposition that the product map of any finite number of continuous maps is continuous by the product topology.
Target Context
- The reader will have a description and a proof of the proposition that some para-product maps of continuous maps are continuous.
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 1
For any finite number of topological spaces, \(T_0\), \(T_{i_1}\), and \(T_{i_2}\) where i = 1, 2, . . ., k, any corresponding number of continuous maps, \(f_i: T_0 \times T_{i_1} \rightarrow T_{i_2}\), the para-product map of the maps, \(f_{k + 1}: T_0 \times T_{1_1} \times T_{2_1} \times . . . \times T_{k_1} \rightarrow T_{1_2} \times T_{2_2} \times . . . \times T_{k_2} = (f_1, f_2, . . ., f_k)\) is continuous.
2: Proof 1
Take \(f_{k + 2}: T_0 \times T_{1_1} \times T_{2_1} \times . . . \times T_{k_1} \rightarrow T_0 \times T_{1_1} \times T_0 \times T_{2_1} \times . . . \times T_0 \times T_{k_1}\), \(f_{k + 2} (p_0, p_1, . . ., p_k) = (p_0, p_1, p_0, p_2, . . ., p_0, p_k)\), which is continuous, because for any \(U \subseteq T_0 \times T_{1_1} \times T_0 \times T_{2_1} \times . . . \times T_0 \times T_{k_1}\), \(U = \cup_i U_{01_i} \times U_{1_i} \times U_{02_i} \times U_{2_i} \times . . . \times U_{0k_i} \times U_{k_i}\) where {i} is a possibly uncountable indexes set and \(U_{0j_i}\) is open on \(T_0\) and \(U_{j_i}\) is open on \(T_{j_1}\), by the product topology, because of a characterization of product topology, \(f_{k + 2}^{-1} (U) = \cup_{i} f_{k + 2}^{-1} (U_{01_i} \times U_{1_i} \times U_{02_i} \times U_{2_i} \times . . . \times U_{0k_i} \times U_{k_i})\), it is suffice to show the openness of each term, \(f_{k + 2}^{-1} (U_{01_i} \times U_{1_i} \times U_{02_i} \times U_{2_i} \times . . . \times U_{0k_i} \times U_{k_i}) = \cap_{j} U_{0j_i} \times U_{2_i} \times U_{3_i} \times . . . \times U_{k_i}\), because for any point, \(p = (p_0, p_1, . . ., p_k) \in f_{k + 2}^{-1} (U_{01_i} \times U_{1_i} \times U_{02_i} \times U_{2_i} \times . . . \times U_{0k_i} \times U_{k_i})\), \(f_{k + 2} (p) \in U_{01_i} \times U_{1_i} \times U_{02_i} \times U_{2_i} \times . . . \times U_{0k_i} \times U_{k_i}\), but \(p_0 \in \cap_{j} U_{0j_i}\), \(p_1 \in U_{1_i}\), \(p_2 \in U_{2_i}\), . . ., \(p_k \in U_{k_i}\), so, \(p = (p_0, p_1, . . ., p_k) \in \cap_{j} U_{0j_i} \times U_{1_i} \times U_{2_i} \times . . . \times U_{k_i}\), for any point, \(p = (p_0, p_1, . . ., p_k) \in \cap_{j} U_{0j_i} \times U_{2_i} \times U_{3_i} \times . . . \times U_{k_i}\), \(f_{k + 2} (p) = (p_0, p_1, p_0, p_2, . . ., p_0, p_k) \in U_{01_i} \times U_{1_i} \times U_{02_i} \times U_{2_i} \times . . . \times U_{0k_i} \times U_{k_i}\), so, \(p \in f_{k + 2}^{-1} (U_{01_i} \times U_{1_i} \times U_{02_i} \times U_{2_i} \times . . . \times U_{0k_i} \times U_{k_i})\), but \(\cap_{j} U_{0j_i} \times U_{1_i} \times U_{2_i} \times . . . \times U_{k_i}\) is open on \(T_0 \times T_{1_1} \times . . . \times T_{k_1}\) by the product topology. Now, for \(\tilde{f_{k + 1}}: T_0 \times T_{1_1} \times T_0 \times T_{2_1} \times . . . \times T_0 \times T_{k_1} \rightarrow T_{1_2} \times T_{2_2} \times . . . \times T_{k_2} = (f_1, f_2, . . ., f_k)\), continuous because of the proposition that the product map of any finite number of continuous maps is continuous by the product topology, \(f_{k + 1} = \tilde{f_{k + 1}} \circ f_{k + 2} \), which is continuous as a compound of continuous maps.
3: Note 1
It is called "para-product map" because it is not really a product map, because \(T_0\) is shared by the component maps.
4: Description 2
For any finite number of topological spaces, \(T_0\) and \(T_{i_2}\) where i = 1, 2, . . ., k, any corresponding number of continuous maps, \(f_i: T_0 \rightarrow T_{i_2}\), the para-product map of the maps, \(f_{k + 1}: T_0 \rightarrow T_{1_2} \times T_{2_2} \times . . . \times T_{k_2} = (f_1, f_2, . . ., f_k)\) is continuous.
5: Proof 2
As \(f_i: T_0 \rightarrow T_{i_2}\) is continuous, \(\tilde{f_i}: T_0 \times \{0\} \rightarrow T_{i_2}\), \(\tilde{f_i} (p, 0) = (f_i (p))\), is continuous, because for any open set, \(U \in T_{i_2}\), \(\tilde{f_i}^{-1} (U) = f_i^{-1} (U) \times \{0\}\), because for any point, \((p, 0) \in \tilde{f_i}^{-1} (U)\), \(\tilde{f_i} (p, 0) = f_i (p) \in U\), so, \(p \in f_i^{-1} (U)\), so, \((p, 0) \in f_i^{-1} (U) \times \{0\}\), and for any point, \((p, 0) \in f_i^{-1} (U) \times \{0\}\), \(\tilde{f_i} (p, 0) = f (p) \in U\), so, \((p, 0) \in \tilde{f_i}^{-1} (U)\), and \(f_i^{-1} (U) \times \{0\}\) is open by the product topology. By Description 1, \(\tilde{f_{k + 1}}: T_0 \times \{0\} \times \{0\} . . . \times \{0\} \rightarrow T_{1_2} \times T_{2_2} \times . . . \times T_{k_2} = (\tilde{f_1}, \tilde{f_2}, . . ., \tilde{f_k})\) is continuous. But \(f_{k + 2}: T_0 \rightarrow T_0 \times \{0\} \times \{0\} . . . \times \{0\}\), \(f_{k + 2} (p) = (p, 0, 0, . . ., 0)\) is continuous because for any open set, \(U \in T_0 \times \{0\} \times \{0\} . . . \times \{0\}\), \(U = U_0 \times \{0\} \times \{0\} . . . \times \{0\}\) where \(U_0\) is open on \(T_0\), by the product topology, and \(f_{k + 2}^{-1} (U) = U_0\). Now, \(f_{k + 1} = \tilde{f_{k + 1}} \circ f_{k + 2}\), a compound of continuous maps, continuous.
References
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