description/proof of that for topological space and its 2 products with Euclidean topological spaces, map between products fiber-preserving and linear on fiber is continuous iff canonical matrix is continuous
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of %field name% matrices space.
- The reader knows a definition of Euclidean topology.
- The reader knows a definition of product topology.
- The reader knows a definition of continuous map.
- The reader admits the proposition that any map from any topological space into any finite product topological space is continuous if and only if all the component maps are continuous.
- The reader admits the proposition that any matrices multiplications map is continuous.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space and its 2 products with any Euclidean topological spaces, any map between the products fiber-preserving and linear on each fiber is continuous if and only if the canonical matrix is 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: Structured Description
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(\mathbb{R}^{d_1}\): \(= \text{ the Euclidean topological space }\)
\(\mathbb{R}^{d_2}\): \(= \text{ the Euclidean topological space }\)
\(T \times \mathbb{R}^{d_1}\): \(= \text{ the product topological space }\)
\(T \times \mathbb{R}^{d_2}\): \(= \text{ the product topological space }\)
\(\mathbb{R}^{d_2 d_1}\): \(= \text{ the Euclidean topological space }\)
\(M\): \(: T \to \{\text{ the } d_2 x d_1 \text{ real matrices }\} \subseteq \mathbb{R}^{d_2 d_1}\), with the codomain as the topological subspace
\(f\): \(: T \times \mathbb{R}^{d_1} \to T \times \mathbb{R}^{d_2}, (t, v) \mapsto (t, M (t) v)\)
//
Statements:
\(f \in \{\text{ the continuous maps }\}\)
\(\iff\)
\(M \in \text{ the continuous maps }\)
//
2: Note
\(T \times \mathbb{R}^{d_1}\) and \(T \times \mathbb{R}^{d_2}\) are canonically regarded as vectors bundles, and "fiber" s are with respect to the vectors bundles.
3: Proof
Whole Strategy: Step 1: suppose that \(f\) is continuous, and see that \(M\) is continuous, according to the definition of continuous map; Step 2: suppose that \(M\) is continuous, and see that \(f\) is continuous, according to the definition of continuous map.
Step 1:
Let us suppose that \(f\) is continuous.
Let us see that \(M\) is continuous.
Let \((t, v) \in T \times \mathbb{R}^{d_1}\) be any.
For any open neighborhood of \(f ((t, v)) = (t, M (t) v) \in T \times \mathbb{R}^{d_2}\), there is a \(U_t \times C_{M (t) v, \epsilon} \subseteq T \times \mathbb{R}^{d_2}\) such that \((t, M (t) v) \in U_t \times C_{M (t) v, \epsilon}\) contained in the neighborhood, where \(U_t \subseteq T\) is an open neighborhood of \(t\) and \(C_{M (t) v, \epsilon} \subseteq \mathbb{R}^{d_2}\) is the open cube around \(M (t) v\).
There is an open neighborhood of \((t, v)\), \(V_t \times V_v \subseteq T \times \mathbb{R}^{d_1}\), such that \(f (V_t \times V_v) \subseteq U_t \times C_{M (t) v, \epsilon}\), where \(V_t \subseteq T\) is an open neighborhood of \(t\) and \(V_v \subseteq \mathbb{R}^{d_1}\) is an open neighborhood of \(v\).
Especially, \(f (V_t \times \{v\}) \subseteq U_t \times C_{M (t) v, \epsilon}\).
For each \(t' \in V_t\), \(f ((t', v)) = (t', M (t') v)\), and \(M (t') v \subseteq C_{M (t) v, \epsilon}\).
Especially, let us take \(v = (0, ..., 0, 1, 0, ..., 0)^t\) where \(1\) is the \(j\)-th component.
Then, \(M (t') v = (M^1_j (t'), ..., M^{d_2}_j (t'))^t \subseteq C_{(M^1_j (t), ..., M^{d_2}_j (t))^t, \epsilon}\), which means that \(M^k_j (t) - \epsilon \lt M^k_j (t') \lt M^k_j (t) + \epsilon\), which means that each \(M^k_j\) is continuous.
That means that \(M\) is continuous, by the proposition that any map from any topological space into any finite product topological space is continuous if and only if all the component maps are continuous.
Step 2:
Let us suppose that \(M\) is continuous.
Let us see that \(f\) is continuous.
Let \((t, v) \in T \times \mathbb{R}^{d_1}\) be any.
For any open neighborhood of \(f ((t, v)) = (t, M (t) v) \in T \times \mathbb{R}^{d_2}\), there is a \(U_t \times C_{M (t) v, \epsilon} \subseteq T \times \mathbb{R}^{d_2}\) such that \((t, M (t) v) \in U_t \times C_{M (t) v, \epsilon}\) contained in the neighborhood, where \(U_t \subseteq T\) is an open neighborhood of \(t\) and \(C_{M (t) v, \epsilon} \subseteq \mathbb{R}^{d_2}\) is an open cube around \(M (t) v\).
As any matrices multiplications maps is continuous by the proposition that any matrices multiplications map is continuous, there are an open cube around \(M (t) \in \mathbb{R}^{d_2 d_1}\), \(C_{M (t), \delta} \subseteq \mathbb{R}^{d_2 d_1}\), and an open cube around \(v \in \mathbb{R}^{d_1}\), \(C_{v, \delta}\), such that the multiplications are contained in \(C_{M (t) v, \epsilon}\).
As \(M\) is continuous, there is an open neighborhood of \(t\), \(V_t \subseteq T\), such that \(M (V_t) \subseteq C_{M (t), \delta}\).
Then, let us think of the open neighborhood of \((t, v)\), \((U_t \cap V_t) \times C_{v, \delta} \subseteq T \times \mathbb{R}^{d_1}\).
\(f ((U_t \cap V_t) \times C_{v, \delta}) \subseteq U_t \times C_{M (t) v, \epsilon}\), because for each \((t', v') \in (U_t \cap V_t) \times C_{v, \delta}\), \(f (t', v') = (t', M (t') v')\), but \(t' \in U_t\) and \(M (t') v' \in C_{M (t) v, \epsilon}\), because \(M (t') \in C_{M (t), \delta}\) and \(v' \in C_{v, \delta}\).
So, \(f\) is continuous.
