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Is every 2 dimensional manifold whose boundary is a cycle, a continuous image of the unit disc? Maybe it happens if the space is good enough? I wanted to prove an equality between two definitions I've seen to simply-connectedness. (Every continuous image the circle is null homotopic, and every cycle is a boundary of another manifold). I also want to know if the vice versa holds: if any continuous image of the unit disc is a manifold. Please keep in mind I don't have a lot of knowledge on the subject. I wasn't sure about which tags I should add to this question. Thanks!

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    $\begingroup$ Robert Israel has answered the question you asked, but I expect it is not exactly the question you wanted to ask. If you remove an open disk from a torus, you are left with a two manifold with a circle for a boundary. It is evidently not simply connected; it is not a topological disk. $\endgroup$ – yasmar Dec 9 '12 at 9:41
  • $\begingroup$ This is kind of a duplicate of math.stackexchange.com/questions/18083/… $\endgroup$ – Jason DeVito Dec 9 '12 at 15:43
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Any compact, connected, locally connected second-countable Hausdorff space (in particular any compact connected manifold with boundary) is a continuous image of the unit interval, and thus of the unit disk. See the Hahn–Mazurkiewicz theorem.

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  • $\begingroup$ Thanks. For the equality of the definitions I also need the other way around: To show that any continuous image of the disc is a manifold. I'm not sure if it's true because of the extra differential structure I'm not sure a continuous function preserves... $\endgroup$ – nodwj Dec 9 '12 at 17:23
  • $\begingroup$ No, of course it isn't true. It's very easy to construct counterexamples. $\endgroup$ – Robert Israel Dec 9 '12 at 21:12
  • $\begingroup$ Then how every image of the circle is null homotopic => every cycle is a boundary? $\endgroup$ – nodwj Dec 10 '12 at 6:52

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