291: Some Para-Product Maps of Continuous Maps Are Continuous

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A description/proof of that some para-product maps of continuous maps are continuous

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Topics

About: topological space
About: map

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The table of contents of this article

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

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Starting Context

• The reader knows a definition of product topology.
• The reader knows a definition of continuous map.
• The reader admits a characterization of product topology.
• The reader admits the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
• The reader admits the proposition that the preimage by any product map is the product of the preimages by the component maps.
• The reader admits the proposition that the product map of any finite number of continuous maps is continuous by the product topology.

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Target Context

• The reader will have a description and a proof of the proposition that some para-product maps of continuous maps are 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: Description 1

For any finite number of topological spaces, $T_0$, $T_{i_1}$, and $T_{i_2}$ where i = 1, 2, . . ., k, any corresponding number of continuous maps, $f_i:T_0\times T_{i_1}\to T_{i_2}$, the para-product map of the maps, $f_{k+1}:T_0\times T_{1_1}\times T_{2_1}\times ...\times T_{k_1}\to T_{1_2}\times T_{2_2}\times ...\times T_{k_2}=(f_1,f_2,...,f_k)$ is continuous.

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2: Proof 1

Take $f_{k+2}:T_0\times T_{1_1}\times T_{2_1}\times ...\times T_{k_1}\to T_0\times T_{1_1}\times T_0\times T_{2_1}\times ...\times T_0\times T_{k_1}$, $f_{k+2}(p_0,p_1,...,p_k)=(p_0,p_1,p_0,p_2,...,p_0,p_k)$, which is continuous, because for any $U\subseteq T_0\times T_{1_1}\times T_0\times T_{2_1}\times ...\times T_0\times T_{k_1}$, $U={\cup }_i{U}_{{01}_i}\times U_{1_i}\times U_{{02}_i}\times U_{2_i}\times ...\times U_{0k_i}\times U_{k_i}$ where {i} is a possibly uncountable indexes set and $U_{0j_i}$ is open on $T_0$ and $U_{j_i}$ is open on $T_{j_1}$, by the product topology, because of a characterization of product topology, $f_{k+2}^{-1}(U)={\cup }_i{f}_{k+2}^{-1}(U_{{01}_i}\times U_{1_i}\times U_{{02}_i}\times U_{2_i}\times ...\times U_{0k_i}\times U_{k_i})$, it is suffice to show the openness of each term, $f_{k+2}^{-1}(U_{{01}_i}\times U_{1_i}\times U_{{02}_i}\times U_{2_i}\times ...\times U_{0k_i}\times U_{k_i})={\cap }_j{U}_{0j_i}\times U_{2_i}\times U_{3_i}\times ...\times U_{k_i}$, because for any point, $p=(p_0,p_1,...,p_k)\in f_{k+2}^{-1}(U_{{01}_i}\times U_{1_i}\times U_{{02}_i}\times U_{2_i}\times ...\times U_{0k_i}\times U_{k_i})$, $f_{k+2}(p)\in U_{{01}_i}\times U_{1_i}\times U_{{02}_i}\times U_{2_i}\times ...\times U_{0k_i}\times U_{k_i}$, but $p_0\in {\cap }_j{U}_{0j_i}$, $p_1\in U_{1_i}$, $p_2\in U_{2_i}$, . . ., $p_k\in U_{k_i}$, so, $p=(p_0,p_1,...,p_k)\in {\cap }_j{U}_{0j_i}\times U_{1_i}\times U_{2_i}\times ...\times U_{k_i}$, for any point, $p=(p_0,p_1,...,p_k)\in {\cap }_j{U}_{0j_i}\times U_{2_i}\times U_{3_i}\times ...\times U_{k_i}$, $f_{k+2}(p)=(p_0,p_1,p_0,p_2,...,p_0,p_k)\in U_{{01}_i}\times U_{1_i}\times U_{{02}_i}\times U_{2_i}\times ...\times U_{0k_i}\times U_{k_i}$, so, $p\in f_{k+2}^{-1}(U_{{01}_i}\times U_{1_i}\times U_{{02}_i}\times U_{2_i}\times ...\times U_{0k_i}\times U_{k_i})$, but ${\cap }_j{U}_{0j_i}\times U_{1_i}\times U_{2_i}\times ...\times U_{k_i}$ is open on $T_0\times T_{1_1}\times ...\times T_{k_1}$ by the product topology. Now, for $\widetilde{f_{k+1}}:T_0\times T_{1_1}\times T_0\times T_{2_1}\times ...\times T_0\times T_{k_1}\to T_{1_2}\times T_{2_2}\times ...\times T_{k_2}=(f_1,f_2,...,f_k)$, continuous because of the proposition that the product map of any finite number of continuous maps is continuous by the product topology, $f_{k+1}=\widetilde{f_{k+1}}\circ f_{k+2}$, which is continuous as a compound of continuous maps.

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3: Note 1

It is called "para-product map" because it is not really a product map, because $T_0$ is shared by the component maps.

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4: Description 2

For any finite number of topological spaces, $T_0$ and $T_{i_2}$ where i = 1, 2, . . ., k, any corresponding number of continuous maps, $f_i:T_0\to T_{i_2}$, the para-product map of the maps, $f_{k+1}:T_0\to T_{1_2}\times T_{2_2}\times ...\times T_{k_2}=(f_1,f_2,...,f_k)$ is continuous.

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5: Proof 2

As $f_i:T_0\to T_{i_2}$ is continuous, $\widetilde{f_i}:T_0\times \{0\}\to T_{i_2}$, $\widetilde{f_i}(p,0)=(f_i(p))$, is continuous, because for any open set, $U\in T_{i_2}$, ${\widetilde{f_i}}^{-1}(U)=f_i^{-1}(U)\times \{0\}$, because for any point, $(p,0)\in {\widetilde{f_i}}^{-1}(U)$, $\widetilde{f_i}(p,0)=f_i(p)\in U$, so, $p\in f_i^{-1}(U)$, so, $(p,0)\in f_i^{-1}(U)\times \{0\}$, and for any point, $(p,0)\in f_i^{-1}(U)\times \{0\}$, $\widetilde{f_i}(p,0)=f(p)\in U$, so, $(p,0)\in {\widetilde{f_i}}^{-1}(U)$, and $f_i^{-1}(U)\times \{0\}$ is open by the product topology. By Description 1, $\widetilde{f_{k+1}}:T_0\times \{0\}\times \{0\}...\times \{0\}\to T_{1_2}\times T_{2_2}\times ...\times T_{k_2}=(\widetilde{f_1},\widetilde{f_2},...,\widetilde{f_k})$ is continuous. But $f_{k+2}:T_0\to T_0\times \{0\}\times \{0\}...\times \{0\}$, $f_{k+2}(p)=(p,0,0,...,0)$ is continuous because for any open set, $U\in T_0\times \{0\}\times \{0\}...\times \{0\}$, $U=U_0\times \{0\}\times \{0\}...\times \{0\}$ where $U_0$ is open on $T_0$, by the product topology, and $f_{k+2}^{-1}(U)=U_0$. Now, $f_{k+1}=\widetilde{f_{k+1}}\circ f_{k+2}$, a compound of continuous maps, continuous.

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References

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