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Give an example of an injective continuous map of $(0,1)$ to $\mathbb{R}^2$ which is not a homeomorphism onto the image of $(0,1)$.

Can anyone help me out here? This is part of a problem sheet on compactness oddly enough; yet, all I know from my notes is that a continuous, bijective map from a compact set is a homeomorphism onto the image of such a compact set.

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    $\begingroup$ Not sure why this has 5 upvotes, would you say it is "well researched"? It shows no effort. Giving hints is something I encourage but upvotes should be reserved for "this person has tried" $\endgroup$ – Alec Teal Jun 7 '15 at 3:03
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HINT: $\qquad\qquad\qquad\qquad$

enter image description here

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  • $\begingroup$ Out of curiosity, would a circle with one point removed also do the job? $\endgroup$ – Clement C. Jun 6 '15 at 17:03
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    $\begingroup$ @Clement: No, that actually is homeomorphic to $(0,1)$. You need to make one of the loose ends approach a point on the image. $\endgroup$ – Brian M. Scott Jun 6 '15 at 17:14
  • $\begingroup$ Ha, I was wondering whether making the images of the two loose ends close to each other was enough (to contradict continuity of the inverse). $\endgroup$ – Clement C. Jun 6 '15 at 17:16
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    $\begingroup$ @Clement: No, as long as there’s an actual gap, even of one point, the inverse will be continuous. $\endgroup$ – Brian M. Scott Jun 6 '15 at 17:17
  • $\begingroup$ I'll think about it to make sure I understand -- thanks! $\endgroup$ – Clement C. Jun 6 '15 at 17:20
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Hint Find a suitable parameterization of the figure eight curve:

enter image description here

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