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

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)

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

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.