A description/proof of characteristic property of product topology
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 any topological spaces map is continuous if and only if for every net on the domain that converges to any point, the composition of the map after the net converges to the image of the point.
- The reader admits the proposition that any net to any product topological space converges to a point if and only if the projection to each constituent space after the net converges to the corresponding component of the point.
Target Context
- The reader will have a description and a proof of the proposition that for any product topological space, any map from any topological space to the product topological space is continuous if and only if the composition of the projection of the product topological space to each constituent space after the map is continuous, and the product topology is the unique topology that makes the product topological space have that property.
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 product topological space, \(T_2 = \times_{\alpha} T_\alpha\) where \(\alpha \in A\) is any possibly uncountable indices set, any map from any topological space, \(f: T_1 \rightarrow T_2\), is continuous if and only if \(\pi_\alpha \circ f: T_1 \rightarrow T_2 \rightarrow T_\alpha\), where \(\pi_\alpha: T_2 \rightarrow T_\alpha\) is the projection map, is continuous. And the product topology is the unique topology that makes the product topological space have that property.
2: Proof
Let us suppose that \(\pi_\alpha \circ f\) is continuous. For any net, \(n: D \rightarrow T_1\), that converges to \(p \in T_1\), \(\pi_\alpha \circ f \circ n\) converges to \(\pi_\alpha \circ f (p)\), by the proposition that any topological spaces map is continuous if and only if for every net on the domain that converges to any point, the composition of the map after the net converges to the image of the point. So, \(f \circ n\) converges to \(f (p)\), by the proposition that any net to any product topological space converges to a point if and only if the projection to each constituent space after the net converges to the corresponding component of the point. So, \(f\) is continuous, by the proposition that any topological spaces map is continuous if and only if for every net on the domain that converges to any point, the composition of the map after the net converges to the image of the point.
Let us suppose that \(f\) is continuous. For any net, \(n: D \rightarrow T_1\), that converges to \(p \in T_1\), \(f \circ n\) converges to \(f (p)\), by the proposition that any topological spaces map is continuous if and only if for every net on the domain that converges to any point, the composition of the map after the net converges to the image of the point. \(\pi_\alpha \circ f \circ n\) converges to \(\pi_\alpha \circ f (p)\), by the proposition that any net to any product topological space converges to a point if and only if the projection to each constituent space after the net converges to the corresponding component of the point. So, \(\pi_\alpha \circ f\) is continuous, by the proposition that any topological spaces map is continuous if and only if for every net on the domain that converges to any point, the composition of the map after the net converges to the image of the point.
Let us suppose that \(T'_2\) has the set of \(T_2\) but has a not-necessarily product topology while it satisfies the property. Let us take \(T_1\) to be \(T'_2\) and \(f\) to be the identity map. Then, \(f\) is continuous and \(\pi_\alpha \circ f\) is continuous. For any open set, \(U_\beta \subseteq T_\beta\), \((\pi_\beta \circ f)^{-1} (U_\beta) = \times_{\alpha} U_\alpha\), where \(U_\alpha = U_\beta\) for \(\alpha = \beta\) and \(U_\alpha = T_\alpha\) for \(\alpha \neq \beta\), is open on \(T'_2\). So, any \(\times_{\alpha} U_\alpha\), where \(U_\alpha = U_\beta\) for some finite \(\alpha\) s and \(U_\alpha = T_\alpha\) for the rest \(\alpha\) s, is open as the intersection of some finite open sets. So, \(T'_2\) has the product topology or a larger topology. Let us take \(T_1\) to be \(T_2\) (with the product topology) and \(f\) to be the identity map. For any open set, \(U_\beta \subseteq T_\beta\), \((\pi_\beta \circ f)^{-1} (U_\beta) = \times_{\alpha} U_\alpha\), where \(U_\alpha = U_\beta\) for \(\alpha = \beta\) and \(U_\alpha = T_\alpha\) for \(\alpha \neq \beta\), is open on \(T_2\). So, \(\pi_\alpha \circ f\) is continuous and \(f\) is continuous. For any open set, \(U \subseteq T'_2\), \(f^{-1} (U) = U\) is open on \(T_2\). So, \(T'_2\) has no larger topology after all.
