Submanifold of a smooth manifold is closed iff the inclusion map is proper

Question

I am studying manifolds and have come across

Let $M$ be a smooth manifold. Show that a submanifold of $M$ is closed in $M$ if and only if the inclusion map is proper.

Answer 1

I will assume you're using the definitions of manifold, submanifold, and proper which appear in Lee's Introduction to Topological Manifolds.

It is clear that a submanifold $S$ is closed iff the inclusion $i\colon S\to M$ is a closed map. First we show that proper implies closed; this holds for arbitrary maps between compactly generated Hausdorff spaces (proven below).

On the other hand, we can show directly that $i$ is proper. Identify $S$ with $i(S)\subset M$. Then $i^{-1}(K)=K\cap S$ for any $K\subset M$.

Suppose that $K$ is compact. Since manifolds are Hausdorff, it follows that $K$ is closed. Since $S$ is closed, it follows that $K\cap S$ is closed. Thus as a closed subset of the compact set $K$, the latter set is compact. This proves that $i$ is proper.

Hence the map is closed if and only if it is proper, so the result follows.

Proof of auxiliary result (proper implies closed for CGH spaces): Let $f\colon X\to Y$ be an arbitrary proper map between CGH spaces and consider any closed $F\subset X$. To show that $f(F)$ is closed, it suffices to show that $f(F)\cap K$ is closed for every compact $K\subset Y$. Since $f$ is proper, $f^{-1}(K)$ is compact. By the closed map lemma, it follows that $f_{f^{-1}(K)}$ is a closed map. Thus $f(F)\cap K=f_{f^{-1}(K)}(F)$ is closed and the result follows.