344: Topological Space Is Connected if Quotient Space and Each Element of Quotient Space Are Connected

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A description/proof of that topological space is connected if quotient space and each element of quotient space are connected

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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

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Starting Context

• The reader knows a definition of quotient topology on set with respect to map.
• The reader knows a definition of connected topological space.
• The reader knows a definition of subspace topology.

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Target Context

• The reader will have a description and a proof of the proposition that any topological space is connected if a quotient space of it and each element of the quotient space are connected.

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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

Any topological space, $T$, is connected if a quotient space, $T/\pi$, where $\pi :T\to T/\pi$, and each element, $p\in T/\pi$, as the subspace of $T$ are connected.

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2: Proof

Let us suppose that $T/\pi$ and each $p$ are connected. Let us suppose that $T$ was not connected. There would be nonempty open sets, $U_1,U_2\subseteq T$, such that $T=U_1\cup U_2$ and $U_1\cap U_2=\mathrm{\varnothing }$. $U_i\cap p$ is open on $p$ by the definition of subspace topology. $p=(U_1\cap p)\cup (U_2\cap p)$, $(U_1\cap p)\cap (U_2\cap p)=\mathrm{\varnothing }$. As $p$ is connected, $U_1\cap p=\mathrm{\varnothing }$ or $U_2\cap p=\mathrm{\varnothing }$. So, $U_i={\cup }_{\alpha \in A_i}{p}_{\alpha }$ where $A_1\cup A_2$ is an indexes set for the elements of $T/\pi$ and $A_1\cap A_2=\mathrm{\varnothing }$. $U_i={\pi }^{-1}(\{p_{\alpha }|\alpha \in A_i\})$, and $\{p_{\alpha }|\alpha \in A_i\}$ is open on $T/\pi$ by the definition of quotient topology. $T/\pi =\{p_{\alpha }\in T/\pi |\alpha \in A_1\}\cup \{p_{\alpha }\in T/\pi |\alpha \in A_2\}$ and $\{p_{\alpha }\in T/\pi |\alpha \in A_1\}\cap \{p_{\alpha }\in T/\pi |\alpha \in A_2\}=\mathrm{\varnothing }$ while $\{p_{\alpha }\in T/\pi |\alpha \in A_i\}\ne \mathrm{\varnothing }$, a contradiction against $T/\pi$'s being connected.

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References

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