description/proof of that for topological space and its 2 products with Euclidean topological spaces, injective continuous map between products fiber-preserving and linear on fiber is continuous embedding
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 Euclidean topological space.
- The reader knows a definition of injection.
- The reader knows a definition of continuous map.
- The reader knows a definition of continuous embedding.
- The reader admits 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.
- The reader admits the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous.
- 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.
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 injective continuous map between the products fiber-preserving and linear on each fiber is a continuous embedding.
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
\(f\): \(: T \times \mathbb{R}^{d_1} \to T \times \mathbb{R}^{d_2}, (t, v) \mapsto (t, M (t) v)\), \(\in \{\text{ the injections }\} \cap \{\text{ the continuous maps }\}\)
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Statements:
\(f \in \{\text{ the continuous embeddings }\}\)
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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: see that \(M\) is continuous; Step 2: see that for each \(t \in T\), \(M (t)\) is rank \(d_1\), and take a \(d_1 x d_1\) invertible submatrix of \(M (t)\); Step 3: take an open neighborhood of \(t\), \(U_t \subseteq T\), on which the submatrix is invertible, and take the open cover of \(T\) with such open subsets; Step 4: take the open cover of \(f (T \times \mathbb{R}^{d_1}) \subseteq T \times \mathbb{R}^{d_2}\) with respect to the open cover of \(T\); Step 5: let \(f': T \times \mathbb{R}^{d_1} \to f (T \times \mathbb{R}^{d_1}) \subseteq T \times \mathbb{R}^{d_2}\) be the codomain restriction of \(f\), and see that \(f'^{-1}\) is continuous on each element of the open cover; Step 6: conclude the proposition.
Step 1:
\(M\) is continuous, by 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.
Step 2:
Let \(t \in T\) be any.
As \(f\) is injective, \(M (t)\) is injective, because it is the restriction of \(f\).
That means that \(M (t)\) is rank \(d_1\).
So, there is a \(d_1 x d_1\) invertible submatrix of \(M (t)\), which means that the determinant of the submatrix is nonzero.
Step 3:
As \(M\) is continuous, the determinant of the submatrix is continuous, so, there is an open neighborhood of \(t\), \(U_t \subseteq T\), on which the determinant is nonzero, which means that the submatrix is invertible on \(U_t\).
Such \(U_t\) s cover \(T\), so, we have the open cover of \(T\), \(\{U_t \vert t \in T\}\).
Step 4:
\(\{U_t \times \mathbb{R}^{d_2} \vert t \in T\}\) is an open cover of \(T \times \mathbb{R}^{d_2}\).
\(\{(U_t \times \mathbb{R}^{d_2}) \cap f (T \times \mathbb{R}^{d_1}) \vert t \in T\}\) is an open cover of \(f (T \times \mathbb{R}^{d_1}) \subseteq T \times \mathbb{R}^{d_2}\) with the subspace topology.
Step 5:
Let \(f': T \times \mathbb{R}^{d_1} \to f (T \times \mathbb{R}^{d_1}) \subseteq T \times \mathbb{R}^{d_2}\) be the codomain restriction of \(f\).
While what we need to see is that \(f'^{-1}\) is continuous, we need to see only that \(f'^{-1} \vert_{(U_t \times \mathbb{R}^{d_2}) \cap f (T \times \mathbb{R}^{d_1})}: (U_t \times \mathbb{R}^{d_2}) \cap f (T \times \mathbb{R}^{d_1}) \to T \times \mathbb{R}^{d_1}\) is continuous, by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous.
Let the invertible \(d_1 x d_1\) submatrix be \(M'\).
For each \((t', w') \in (U_t \times \mathbb{R}^{d_2}) \cap f (T \times \mathbb{R}^{d_1})\), let \(f'^{-1} ((t', w')) := (t', v')\). \((t', v') \in U_t \times \mathbb{R}^{d_1}\) and \(f' ((t', v')) = (t', M (t') v' = w')\).
\(M' (t') v' = (w'^{j_1}, ..., w'^{j_{d_1}})^t := w''\), where \(\{j_1, ..., j_{d_1}\} \subseteq \{1, ..., d_2\}\).
\(f'^{-1} ((t', w')) = (t', v') = (t', M' (t')^{-1} w'')\).
As \(M\) is continuous, each component of \(M\) 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, and so, each component of \(M'^{-1}\) is continuous (it is a polynomial of the components of \(M\) divided by the determinant of \(M'\), which does not become zero), and \(M'^{-1}\) is continuous.
By 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, the map, \(g: U_t \times \mathbb{R}^{d_1} \to U_t \times \mathbb{R}^{d_1}, (t', w'') \mapsto (t', M'^{-1} w'')\), is continuous.
For each open neighborhood of \((t', v') \in U_t \times \mathbb{R}^{d_1}\), there is a \(V_{t'} \times V_{v'} \subseteq U_t \times \mathbb{R}^{d_1}\) that is contained in the neighborhood, where \(V_{t'} \subseteq U_t\) is an open neighborhood of \(t'\) and \(V_{v'} \subseteq \mathbb{R}^{d_1}\) is an open neighborhood of \(v'\).
As \(g\) is continuous, there is a \(U_{t'} \times U_{w''} \subseteq U_t \times \mathbb{R}^{d_1}\) such that \(g (U_{t'} \times U_{w''}) \subseteq V_{t'} \times V_{v'}\), where \(U_{t'} \subseteq U_t\) is an open neighborhood of \(t'\) and \(U_{w''} \subseteq \mathbb{R}^{d_1}\) is an open neighborhood of \(w''\).
\(U_{t'} \times U_{w''} \times \mathbb{R}^{d_2 - d_1} \subseteq U_t \times \mathbb{R}^{d_2}\) is an open neighborhood of \((t', w')\), and so, \((U_{t'} \times U_{w''} \times \mathbb{R}^{d_2 - d_1}) \cap f (T \times \mathbb{R}^{d_1}) \subseteq U_t \times \mathbb{R}^{d_2} \cap f (T \times \mathbb{R}^{d_1})\) is an open neighborhood of \((t', w')\).
Then, \(f'^{-1} \vert_{(U_t \times \mathbb{R}^{d_2}) \cap f (T \times \mathbb{R}^{d_1})} ((U_{t'} \times U_{w''} \times \mathbb{R}^{d_2 - d_1}) \cap f (T \times \mathbb{R}^{d_1})) \subseteq V_{t'} \times V_{v'}\), because \(f'^{-1} ((t', w')) = (t', M' (t')^{-1} w'') = g ((t', w''))\) and \(g (U_{t'} \times U_{w''}) \subseteq V_{t'} \times V_{v'}\).
So, \(f'^{-1} \vert_{(U_t \times \mathbb{R}^{d_2}) \cap f (T \times \mathbb{R}^{d_1})}\) is continuous.
Step 6:
So, \(f'^{-1}\) is continuous, and so, \(f'\) is homeomorphic.
So, \(f\) is a continuous embedding.
