Embedding $D^k \to S^n$ is a closed map?

Question

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)

@LordSharktheUnknown Yes, $D^k$ is the $k$-dimensional closed unit disk. — blancket, Nov 28 '19 at 11:49
have a look at this question and it's answer: https://math.stackexchange.com/questions/1327685/continuous-function-from-a-compact-space-to-a-hausdorff-space-is-a-closed-functi — Thomas, Nov 28 '19 at 11:53
The image of a compact set under a continuous map is compact. — Angina Seng, Nov 28 '19 at 11:56
Note that compact sets are closed only in a Hausdorff setting. — k76u4vkweek547v7, Nov 28 '19 at 13:20

score 0 · Accepted Answer · answered Nov 28 '19 at 18:31

0

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.

answered Nov 28 '19 at 18:31

Paul Frost

1 Answers1