Image of Embedding on a Manifold is a Submanifold

Question

While proving this result the author asserts that it is enough to prove that the embedding is actually an open map.

I don't really see why it's actually enough. I think it's linked with the definition of Manifolds which says that They are locally diffeomorphic to open subsets of Euclidean spaces.

EDIT:

What I thought is that suppose $(\phi, U)$ be the local parametrization for $X $then if $f $is open then every open set in $f(X)$ will be of the form$ f(A) $for some open set $A $in $X $. In that case$ (fo\phi, U) $will be the desired local parametrization for $f(X)$. Hence a Manifold.

kindly Help! what am i missing?

thanks & regards

Answer 1

Recall that given two smooth manifolds $M$ and $N$, we say a smooth map $f: M \to N$ is an embedding if it is an immersion which is also a homeomorphism onto its image $f(M)$, with respect to the subspace topology. However, if you know that the map $f$ is an injective immersion and is also an open map onto its image, you can conclude (using the definition of subspace topology) that $f$ is a homeomorphism onto its image.

So, the reason for claiming its enough to prove $f$ is an open map onto its image has nothing to do with the manifold properties; it is merely a topological argument.

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So, writing things out explicitly, here's what you need to prove (its a purely topological argument):

Theorem:

Let $M,N$ be topological spaces, and let $f:M \to N$ be continuous and injective. Equip $f(M)$ with the subspace topology, and further assume that for every open set $U$ of $M$, $f(U)$ is open relative to $f(M)$ (this is what it means for $f$ to be an open map onto its image with the subspace topology). Then, $f: M \to f(M)$ is a homeomorphism.

So really what you have to prove is that the inverse $f^{-1}: f(M) \to M$ (which exists since $f$ is injective) is also continuous. The proof of this is really just 2 lines if you understood the definition of continuity and openness.