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An injective immersion from a compact manifold is an embedding

Statement

Let $M$ be compact and let $N$ be Hausdorff. Every injective smooth immersion $F : M \to N$ is a smooth embedding.

Facts & Assumptions

Given: A compact manifold $M$ , a Hausdorff manifold $N$ , and an injective smooth immersion $F : M \to N$ .

[F1]

A smooth embedding is an injective immersion and a homeomorphism onto its image with the subspace topology (Smooth embeddings).

[L1]

Smooth maps are continuous (Smooth maps are continuous).

[L2]

Closed subsets of compact spaces are compact (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).

[L3]

Compact subsets of Hausdorff spaces are closed (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

[L4]

Hausdorffness is hereditary to subspaces ( $T_{0}$ , $T_{1}$ , and Hausdorffness are hereditary).

Proof

technique · direct

1.1L1L4given

By [L1], $F$ is continuous. Since it is injective, the corestriction $f_{0} : M \to F (M)$ is a continuous bijection. By [L4], the subspace $F (M) \subseteq N$ is Hausdorff.

2.1L1L2L3step 1.1

Let $C \subseteq M$ be closed. Since $M$ is compact, [L2] makes $C$ compact. To show that $f_{0} [C]$ is compact in $F (M)$ , let $U$ be an open cover of $f_{0} [C]$ in the subspace $F (M)$ ; then ${f_{0}^{- 1} (U) : U \in U}$ is an open cover of $C$ , so finitely many members cover $C$ . Applying $f_{0}$ back shows that the same finite subfamily covers $f_{0} [C]$ . Thus $f_{0} [C]$ is compact, hence closed in the Hausdorff space $F (M)$ by [L3].

3.1F1step 1.1step 2.1∎

Step 2.1 shows that $f_{0}$ is a closed bijection. Therefore for every open set $O \subseteq M$ , the complement $M ∖ O$ is closed and $f_{0} [O] = F (M) ∖ f_{0} [M ∖ O]$ is open in $F (M)$ . So $f_{0}$ is an open bijection, hence a homeomorphism. Since $F$ is an injective immersion by hypothesis, [F1] now gives that $F$ is a smooth embedding.

Depends on

Used by

Dependency tree · two levels

28 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John M. Lee, Introduction to Smooth Manifolds, Embeddings (standard reference, not scraped)
Will J. Merry, Differential Geometry, Proposition 6.3 (standard reference, not scraped)