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:

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

  • 1
    $\begingroup$ $\dim F(N)=\dim N$ 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 $\mathbb{R}^n$. $\endgroup$ – freakish Jul 20 '19 at 11:17
  • $\begingroup$ @freakish Oh wait I got it. $\dim N = \dim F(N)$ because of this ? Thanks! $\endgroup$ – user636532 Jul 20 '19 at 12:34
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    $\begingroup$ the link seems overly complicated, but yes. $\endgroup$ – freakish Jul 20 '19 at 12:52
  • $\begingroup$ @freakish Another problem is (1) in Part 2. May you please help? $\endgroup$ – user636532 Jul 21 '19 at 6:48
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    $\begingroup$ 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. $\endgroup$ – freakish Jul 21 '19 at 7:45

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