Suppose $\phi: [0,1] \to \mathbb R^n$ is a continuous map. Does there exist continuous, injective $\psi : [0,1] \to \mathbb R^n$ such that $\psi(0) = \phi(0)$ and $\psi(1) = \phi(1)$ and $Im(\psi) \subset Im(\phi)$?
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Just take $\psi$ to be a straight line segment – cpiegore Feb 25 '16 at 21:25
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Do you want strict inclusion of the images? Otherwise we could just use $\psi = \phi$. – Daron Feb 25 '16 at 21:26
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@Daron What if $\phi$ is not injective? – cpiegore Feb 25 '16 at 21:28
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Whoops! Wasn't paying enough attention. – Daron Feb 25 '16 at 21:29
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@cpiegore: That might not be contained in the image of $\phi$. – Eric Wofsey Feb 25 '16 at 21:29
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@EricWofsey Do you have a specific example where a line segment does not work? – cpiegore Feb 25 '16 at 21:37
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@cpiegore A circular arc, for example. – nicomezi Feb 25 '16 at 21:38