A description/proof of that product map of continuous maps is continuous
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of product topology.
- The reader knows a definition of continuous map.
- 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.
Target Context
- The reader will have a description and a proof of the proposition that the product map of any finite number of continuous maps is continuous by the product topologies.
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 topological spaces, \(T_{1, i}\) and \(T_{2, i}\) where \(i = 1, 2, . . ., k\), any corresponding number of continuous maps, \(f_i: T_{1, i} \rightarrow T_{2, i}\), the product map of the maps, \(f_{k + 1}: T_{1, 1} \times T_{1, 2} \times . . . \times T_{1, k} \rightarrow T_{2, 1} \times T_{2, 2} \times . . . \times T_{2, k} = (f_1, f_2, . . ., f_k)\) is continuous.
2: Proof
For any open set, \(U \in T_{2, 1} \times T_{2, 2} \times . . . \times T_{2, k}\), \(U = \cup_{\alpha \in A} U_{1, \alpha} \times U_{2, \alpha} \times . . . \times U_{k, \alpha}\) where \(A\) is a possibly uncountable indexes set and \(U_{j, \alpha} \subseteq T_{2, j}\) is open on \(T_{2, j}\), by the product topology, because of Note in the article of the definition of product topology. \(f_{k + 1}^{-1} (U) = \cup_{\alpha \in A} f_{k + 1}^{-1} (U_{1, \alpha} \times U_{2, \alpha} \times . . . \times U_{k, \alpha})\) by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets. It is suffice to show the openness of \(f_{k + 1}^{-1} (U_{1, \alpha} \times U_{2, \alpha} \times . . . \times U_{k, \alpha})\). \(f_{k + 1}^{-1} (U_{1, \alpha} \times U_{2, \alpha} \times . . . \times U_{k, \alpha}) = f_1^{-1} (U_{1, \alpha}) \times f_2^{-1} (U_{2, \alpha}) \times . . . \times f_k^{-1} (U_{k, \alpha})\) by the proposition that the preimage by any product map is the product of the preimages by the component maps. \(f_j^{-1} (U_{j, \alpha})\) is open on \(T_{1, j}\) as \(f_j\) is continuous, and the product of them is open on \(T_{1, 1} \times T_{1, 2} \times . . . \times T_{1, k}\) by the product topology.
