838: 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

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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

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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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.

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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.

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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.

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Main Body

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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
$f$: $:T\times {\mathbb{R}}^{d_1}\to T\times {\mathbb{R}}^{d_2},(t,v)\mapsto (t,M(t)v)$
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Statements:
$f\in \{\text{ the continuous maps }\}$
$⟺$
$M\in \text{ the continuous maps }$
//

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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.

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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 $U_t\times C_{M(t)v,ϵ}\subseteq T\times {\mathbb{R}}^{d_2}$ such that $(t,M(t)v)\in U_t\times C_{M(t)v,ϵ}$ contained in the neighborhood, where $U_t\subseteq T$ is an open neighborhood of $t$ and $C_{M(t)v,ϵ}\subseteq {\mathbb{R}}^{d_2}$ 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 $f(V_t\times V_v)\subseteq U_t\times C_{M(t)v,ϵ}$, 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, $f(V_t\times \{v\})\subseteq U_t\times C_{M(t)v,ϵ}$.

For each $t^{\prime }\in V_t$, $f((t^{\prime },v))=(t^{\prime },M(t^{\prime })v)$, and $M(t^{\prime })v\subseteq C_{M(t)v,ϵ}$.

Especially, let us take $v=(0,...,0,1,0,...,0{)}^t$ where $1$ is the $j$-th component.

Then, $M(t^{\prime })v=(M_j^1(t^{\prime }),...,M_j^{d_2}(t^{\prime }){)}^t\subseteq C_{(M_j^1(t),...,M_j^{d_2}(t){)}^t,ϵ}$, which means that $M_j^k(t)-ϵ<M_j^k(t^{\prime })<M_j^k(t)+ϵ$, which means that each $M_j^k$ 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 $U_t\times C_{M(t)v,ϵ}\subseteq T\times {\mathbb{R}}^{d_2}$ such that $(t,M(t)v)\in U_t\times C_{M(t)v,ϵ}$ contained in the neighborhood, where $U_t\subseteq T$ is an open neighborhood of $t$ and $C_{M(t)v,ϵ}\subseteq {\mathbb{R}}^{d_2}$ 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 $C_{M(t)v,ϵ}$.

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 $(t,v)$, $(U_t\cap V_t)\times C_{v,\delta }\subseteq T\times {\mathbb{R}}^{d_1}$.

$f((U_t\cap V_t)\times C_{v,\delta })\subseteq U_t\times C_{M(t)v,ϵ}$, because for each $(t^{\prime },v^{\prime })\in (U_t\cap V_t)\times C_{v,\delta }$, $f(t^{\prime },v^{\prime })=(t^{\prime },M(t^{\prime })v^{\prime })$, but $t^{\prime }\in U_t$ and $M(t^{\prime })v^{\prime }\in C_{M(t)v,ϵ}$, because $M(t^{\prime })\in C_{M(t),\delta }$ and $v^{\prime }\in C_{v,\delta }$.

So, $f$ is continuous.

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References

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