Let $k,n$ be positive integers, and suppose $h:D^k \to S^n$ is an embedding. Then is the image $h(D^k)$ closed in $S^n$?

I know that embedding is not in general a closed map, but in this special case, it seems true, but I'm not sure.

I thought about this question while reading the proof of the Jordan-Brouwer separation theorem (Proposition 2B.1 in Hatcher)

  • $\begingroup$ Can you remind us what $D^k$ is please. $\endgroup$ – Lord Shark the Unknown Nov 28 '19 at 11:48
  • $\begingroup$ @LordSharktheUnknown Yes, $D^k$ is the $k$-dimensional closed unit disk. $\endgroup$ – Quadr Nov 28 '19 at 11:49
  • $\begingroup$ have a look at this question and it's answer: math.stackexchange.com/questions/1327685/… $\endgroup$ – Thomas Nov 28 '19 at 11:53
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    $\begingroup$ The image of a compact set under a continuous map is compact. $\endgroup$ – Lord Shark the Unknown Nov 28 '19 at 11:56
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    $\begingroup$ Oh my god, it was too trivial $\endgroup$ – Quadr Nov 28 '19 at 12:03

This community wiki solution is intended to clear the question from the unanswered queue.

If $f : X \to Y$ is a continuous map and $C \subset X$ is compact, then $f(C)$ is compact. Thus if $Y$ is Hausdorff, then $f(C)$ is closed.


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