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Let $f : \mathbb{R} \rightarrow \mathbb{R}^2$ be a continuous, injective function that is unbounded on both $(-\infty, 0)$ and $(0, \infty)$. Is it possible for the complement of the image of the function to be path-connected?

In this similar question the answer gives a example for which the complement of the image is connected, but not path-connected. Also, there are suggestions in the comments to extend the range of the function to $S^2$ and apply some form of the Jordan curve theorem, which would give a negative answer to this question, but I'm not sure if such extension can work without requiring stronger properties than just unboundedness.

Alex Ravsky
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FelixMP
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    An extension of $f$ to a map $F : S^1 \to S^2$ is possible iff $f(t) \to \infty$ as $t \to \pm \infty$. However, in general the latter is not true. See the example in your link. – Paul Frost May 05 '20 at 05:52

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