# Equivalent definitions for smooth embedding?

Let $$N$$ and $$M$$ be smooth manifolds of respective dimensions $$n$$ and $$m$$. Let $$F:N \to M$$ be a smooth map.

Please verify my proof of the equivalence of the following 2 definitions.

• From An Introduction to Manifolds by Loring W. Tu: Definition 1: Immersion and topological embedding

• From Calculus to Cohomology by Ib Madsen and Jørgen Tornehave: Definition 2: Image is smooth (regular) submanifold (and thus manifold) and diffeomorphism onto image

To prove Definition 1 implies Definition 2:

• Image is smooth submanifold: Tu Theorem 11.13

• Diffeomorphism onto image: Let $$i: F(N) \to M$$ be inclusion. Then the restriction $$\tilde F: N \to F(N)$$, which satisfies $$F = i \circ \tilde F$$ is smooth since $$F$$ smooth by Tu Theorem 11.15. $$\tilde F$$ is also a diffeomorphism through these steps:

• Step 1: $$\tilde F$$, like the original $$F$$, is an immersion.

• Step 2: $$N$$ and $$F(N)$$ have the same dimension.

• Step 3: $$\tilde F$$ is a local diffeomorphism, i.e. $$F$$ is a local diffeomorphism onto its image.

• This follows by Steps 1 and 2 because immersions of manifolds of the same dimension are local diffeomorphisms, as proven here.
• Step 4: $$\tilde F$$ is a diffeomorphism

• This follows by Step 3 because any smooth map $$G$$ of manifolds (with dimensions) is a diffeomorphism if and only if $$G$$ is a bijective local diffeomorphism, as proven here and here.

To prove Definition 2 implies Definition 1:

• Homeomorphism onto image: Diffeomorphism onto image implies homeomorphism onto image, i.e. $$\tilde F$$ diffeomorphism implies $$\tilde F$$ homeomorphism.

• Immersion: Diffeomorphism onto image implies $$\tilde F$$ is immersion. Then, $$F$$ is also an immersion by this again: $\tilde F$ immersion is equivalent to $F$ immersion

• $\dim F(N)=\dim N$ simply from the definition: take any point $y\in F(N)$, consider an open neighbourhood $U$ of $F^{-1}(y)$ homeomorphic to $\mathbb{R}^n$. Then $F(U)$ is an open neighbourhood of $y$ homeomorphic to $\mathbb{R}^n$. – freakish Jul 20 '19 at 11:17
• @freakish Oh wait I got it. $\dim N = \dim F(N)$ because of this ? Thanks! – user636532 Jul 20 '19 at 12:34
• the link seems overly complicated, but yes. – freakish Jul 20 '19 at 12:52
• @freakish Another problem is (1) in Part 2. May you please help? – user636532 Jul 21 '19 at 6:48
• What is your definition of submanifold? One simple approach is that if $F(N)$ inherits its differntial structure from $M$ then obviously $D_p(F)=D_p(\tilde F)$ for any point $p\in N$ and so $\tilde F$ is immersion iff $F$ is. – freakish Jul 21 '19 at 7:45