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.

• 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
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.