439: Continuous Image of Path-Connected Subspace of Domain Is Path-Connected on Codomain

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A description/proof of that continuous image of path-connected subspace of domain is path-connected on codomain

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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 path-connected topological space.
• The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.

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Target Context

• The reader will have a description and a proof of the proposition that any continuous map image of any path-connected subspace of the domain is path-connected on the codomain.

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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 continuous map, $f:T_1\to T_2$, and any path-connected subspace, $T_3\subseteq T_1$, the image, $f(T_3)$, is path-connected on $T_2$.

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2: Proof

For any points, $p_1,p_2\in f(T_3)$, is there a path, $\lambda :[0,r]\to f(T_3)$ such that $p_1=\lambda (0)$ and $p_2=\lambda (r)$? We can take any point from $f^{-1}(p_i)\cap T_3$ as $p_i^{\prime }$. As $T_3$ is path-connected, there is a path, ${\lambda }^{\prime }:[0,1]\to T_3$, such that ${\lambda }^{\prime }(0)=p_1^{\prime }$ and ${\lambda }^{\prime }(1)=p_2^{\prime }$. Let us define $\lambda :=f{|}_{T_3}\circ {\lambda }^{\prime }:[0,1]\to T_3\to f(T_3)$, which is continuous as a composition of continuous maps ($f{|}_{T_3}$ is continuous by the proposition that any restriction of any continuous map on the domain and the codomain is continuous), and $\lambda (0)=p_1$ and $\lambda (1)=p_2$.

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3: Note

If the domain is a path-connected topological space, the domain is a path-connected subspace of itself, so, the image of any path-connected topological space is path-connected.

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References

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