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Thread: pathwise connected subset

  1. #1
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    pathwise connected subset

    Hi,

    Could someone help me prove the following:

    Let $\displaystyle f:X \to Y$ be a continuos map between metric-spaces $\displaystyle (X,d_{X})$ and $\displaystyle (Y,d_{Y})$. Prove that if $\displaystyle T \subseteq X$ is a pathwise connected subset in $\displaystyle X$, then the image $\displaystyle f(T) \subseteq Y$ under $\displaystyle f$ is a pathwise connected subset in $\displaystyle Y$.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    $\displaystyle f:T\rightarrow f(T)$ is surjective so, for $\displaystyle a,b\in f(T)$ there exist $\displaystyle a',b'\in T$ such that $\displaystyle f(a')=a$ and $\displaystyle f(b')=b$ . As $\displaystyle T$ is pathwise connected, there exists a path $\displaystyle \gamma$ from $\displaystyle a'$ to $\displaystyle b'$ contained in $\displaystyle T$. Then, $\displaystyle f\circ \gamma$ is a path from $\displaystyle a$ to $\displaystyle b$ contained in $\displaystyle f(T)$ .


    Fernando Revilla
    Last edited by FernandoRevilla; Jan 12th 2011 at 04:08 AM.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by FernandoRevilla View Post
    $\displaystyle f:T\rightarrow f(T)$ is surjective
    By the way surjective, I suppose you understand why $\displaystyle f:T\rightarrow f(T)$ is surjective .
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