373: Induced Functional Structure on Continuous Topological Spaces Map Codomain Is Functional Structure

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A description/proof of that induced functional structure on continuous topological spaces map codomain is functional structure

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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: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of functional structure on topological space.
• The reader knows a definition of induced functional structure on continuous topological spaces map codomain.

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Target Context

• The reader will have a description and a proof of the proposition that the induced functional structure on the codomain of any continuous topological spaces map is a functional structure.

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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: Description

For any topological spaces, $T_1,T_2$, any functional structure, $\{F_{T_1}(U)\}$, and any continuous map, $\varphi :T_1\to T_2$, the induced functional structure, $\{F_{T_2}(U)=\{f:U\to \mathbb{R}|f\text{ is continuous }\wedge f\circ \varphi {|}_{{\varphi }^{-1}(U)}\in F_{T_1}({\varphi }^{-1}(U))\}\}$, is a functional structure.

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2: Proof

Let us prove that the "induced functional structure" satisfies the 1st condition: $F_{T_2}(U)$ is a sub-algebra of the all-the-continuous functions algebra. $F_{T_2}(U)$ is a subset of the all-the-continuous functions algebra, because $f$ is continuous by the definition. Let us check that the subset is closed under addition. Let us suppose that $f_i\in F_{T_2}(U)$. $f_1+f_2\in F_{T_2}(U)$? $f_1+f_2$ is continuous. $(f_1+f_2)\circ \varphi {|}_{{\varphi }^{-1}(U)}\in F_{T_1}({\varphi }^{-1}(U))$? $f_i\circ \varphi {|}_{{\varphi }^{-1}(U)}\in F_{T_1}({\varphi }^{-1}(U))$. $f_1\circ \varphi +f_2\circ \varphi \in F_{T_1}({\varphi }^{-1}(U))$, but $f_1\circ \varphi +f_2\circ \varphi =(f_1+f_2)\circ \varphi$. Let us check that the subset is closed under scalar multiplications. Let us suppose that $f\in F_{T_2}(U)$ and $r\in \mathbb{R}$. $rf\in F_{T_2}(U)$? $rf$ is continuous. $(rf)\circ \varphi {|}_{{\varphi }^{-1}(U)}\in F_{T_1}({\varphi }^{-1}(U))$? $f\circ \varphi {|}_{{\varphi }^{-1}(U)}\in F_{T_1}({\varphi }^{-1}(U))$. $r(f\circ \varphi ){|}_{{\varphi }^{-1}(U)}\in F_{T_1}({\varphi }^{-1}(U))$, but $r(f\circ \varphi )=(rf)\circ \varphi$. Let us check that the subset is closed under multiplications. Let us suppose that $f_i\in F_{T_2}(U)$. $f_1{f}_2\in F_{T_2}(U)$? $f_1{f}_2$ is continuous. $(f_1{f}_2)\circ \varphi {|}_{{\varphi }^{-1}(U)}\in F_{T_1}({\varphi }^{-1}(U))$? $f_i\circ \varphi {|}_{{\varphi }^{-1}(U)}\in F_{T_1}({\varphi }^{-1}(U))$. $(f_1\circ \varphi )(f_2\circ \varphi ){|}_{{\varphi }^{-1}(U)}\in F_{T_1}({\varphi }^{-1}(U))$, but $(f_1\circ \varphi )(f_2\circ \varphi )=(f_1{f}_2)\circ \varphi$. For any $f_i\in F_{T_2}(U)$, the left distributability, $(f_1+f_2)f_3=f_1{f}_3+f_2{f}_3$, holds. For any $f_i\in F_{T_2}(U)$, the right distributability, $f_3(f_1+f_2)=f_3{f}_1+f_3{f}_1$, holds. For any $f_i\in F_{T_2}(U)$ and $r_i\in \mathbb{R}$, the compatibility with scalars, $(r_1{f}_1)(r_2{f}_2)=(r_1{r}_2)(f_1{f}_2)$, holds.

Let us prove that the "induced functional structure" satisfies the 2nd condition: $F_{T_2}(U)$ contains all the constant functions. For any constant function, $f:U\to \mathbb{R}$, $f\circ \varphi$ is constant, so, $f\circ \varphi \in F_{T_1}({\varphi }^{-1}(U))$.

Let us prove that the "induced functional structure" satisfies the 3rd condition: for any open $V\subseteq U$ and any $f\in F_{T_2}(U)$, $f{|}_V\in F_{T_2}(V)$. $f{|}_V\circ \varphi {|}_{{\varphi }^{-1}(V)}\in F_{T_1}({\varphi }^{-1}(V))$? $f\circ \varphi \in F_{T_1}({\varphi }^{-1}(U))$. ${\varphi }^{-1}(V)\subseteq {\varphi }^{-1}(U)$. $(f\circ \varphi ){|}_{{\varphi }^{-1}(V)}\in F_{T_1}({\varphi }^{-1}(V))$, but $(f\circ \varphi ){|}_{{\varphi }^{-1}(V)}=f{|}_V\circ \varphi {|}_{{\varphi }^{-1}(V)}$, because for any $p\in {\varphi }^{-1}(V)$, $(f\circ \varphi ){|}_{{\varphi }^{-1}(V)}(p)=f{|}_V\circ \varphi (p)$.

Let us prove that the "induced functional structure" satisfies the 4th condition: for any open cover $U={\cup }_{\alpha }{U}_{\alpha }$ and any continuous $f:U\to \mathbb{R}$ such that $f{|}_{U_{\alpha }}\in F_{T_2}(U_{\alpha })$, $f\in F_{T_2}(U)$. $f\circ \varphi {|}_{{\varphi }^{-1}(U)}\in F_{T_1}({\varphi }^{-1}(U))$? $f{|}_{U_{\alpha }}\circ \varphi {|}_{{\varphi }^{-1}(U_{\alpha })}\in F_{T_1}({\varphi }^{-1}(U_{\alpha }))$, but $f{|}_{U_{\alpha }}\circ \varphi {|}_{{\varphi }^{-1}(U_{\alpha })}=(f\circ \varphi ){|}_{{\varphi }^{-1}(U_{\alpha })}$, so, $(f\circ \varphi ){|}_{{\varphi }^{-1}(U_{\alpha })}\in F_{T_1}({\varphi }^{-1}(U_{\alpha }))$. ${\varphi }^{-1}(U)={\cup }_{\alpha }{\varphi }^{-1}(U_{\alpha })$. So, $(f\circ \varphi ){|}_{{\varphi }^{-1}(U)}\in F_{T_1}({\varphi }^{-1}(U))$, but $(f\circ \varphi ){|}_{{\varphi }^{-1}(U)}=f\circ \varphi {|}_{{\varphi }^{-1}(U)}$, so, $f\circ \varphi {|}_{{\varphi }^{-1}(U)}\in F_{T_1}({\varphi }^{-1}(U))$.

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3: Note

Just calling $\{F_{T_2}(U)=\{f:U\to \mathbb{R}|f\text{ is continuous }\wedge f\circ \varphi \in F_{T_1}({\varphi }^{-1}(U))\}\}$ "induced functional structure" does not guarantee that $\{F_{T_2}(U)\}$ is a functional structure, which requires satisfying the 4 conditions.

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References

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