Embedded submanifold and isomorphisms of the ambient space.

Pardon if this is an duplicate question. Say, for instance that I an embedded submanifold $$N$$ of a manifold $$M$$. I also know that $$M$$ is isomorphic (diffeomorphic?) to $$M'$$ via the isomorphism $$f$$. The embedded submanifold $$N$$ is defined via the regular level set theorem (see for instance Lee, ISM first edition, Corr. 8.10). Can I say that $$N$$ also is embedded in $$M'$$?

My attempt: We know that there is an immersion $$F:N \to M$$, and that this immersion is a topological embedding. It feels as if we could consider $$f \circ F : N \to M'$$, and while I see that this is an immersion, I fail to see that it is also a topological embedding. Is this not the way to go? Are there other things I can say?

• The map $f \circ F$ send $N$ homeomorphically onto $f(F(N))$ since $N$ homeomorphic with $F(N)$ through $F$ and $f$, in particular, is a homeomorphism. – Kelvin Lois Oct 1 '20 at 15:14
• Thank you for your answer! So, perhaps a stupid question, how does this give that it is a topological embedding? And also, the embedding into $M$ induces a structure on $N$, what will happen with that in $M'$? Can I say that it still will be from the regular level set theorem? – ejk Oct 1 '20 at 19:29

The key fact if you need, if you want to show that $$f\circ F$$ is a topological embedding is, that restrictions of continous maps are continous:

If $$f:X\to Y$$ is a continous map between topological spaces, $$A\subseteq X$$, $$B\subseteq f(A)$$, then the restriction $$f :A\to B$$ is also continous, where $$A$$ and $$B$$ are equipped with the subspace topologies.

From this it follows that if $$f:X\to Y$$ is a homeomorphism and $$A\subseteq X$$, the restriction $$f:A\to f(A)$$ is also a homeomorphism.

Now in your case you want to show that $$f\circ F:N\to M'$$ is a topological embedding, which means that the restriction $$f\circ F:N\to f(F(N))$$ is a homemorphism.

But this map can be written as the composition

$$N\stackrel{F}\to F(N)\stackrel{f}\to f(F(N))$$

which by the above and the assumption that $$F:N\to M$$ is a topological embedding is a composition of homeomorphisms and hence a homeomorphism.