Are these equivalent for regular submanifold? Open, image of local diffeomorphism, image of injective local diffeomorphism

differential-geometrydifferential-topologygeneral-topologymanifoldssubmanifold

I was thinking about

  1. this: Is open (topological) smooth embedding equivalent to injective local (homeomorphism) diffeomorphism?

  2. how the image of any local diffeomorphism between smooth $m$-manifolds is open

  3. the equivalence of the definitions of smooth regular $k$-submanifold and smooth embedded $k$-submanifold (each of a smooth $m$-manifold)

  4. (not sure about this, but…) $P$ is an immersed submanifold of the smooth $m$-manifold $M$ if and only if $P$ is the image of an injective immersion into $M$, i.e. there exists smooth $x$-manifold $X$ and smooth injective immersion $F:X \to M$ with image($F$)=$F(X)=P$

    • Edit: Oh right. We can't use inclusion map $i: P \to M$ as the injective immersion because it doesn't even make sense to '$i: P \to M$' is smooth because $P$ is not necessarily a smooth manifold! But anyway, open subsets are regular/embedded submanifolds, sooo…

This leads me to conjecture the equivalence of the following for a smooth $m$-manifold $N$ and a subset $A$ of $N$:

  1. $A$ is open in $N$ (if and only if $\iota: A \to N$ is open map; if and only if $A$ is smooth regular/embedded $k$-submanifold of $N$, with $k=n$)
  2. $A$ is the image of a local diffeomorphism into $N$, i.e. there exists smooth $y$-manifold $Y$ and smooth local diffeomorphism $G:Y \to N$ with image($G$)=$G(Y)=A$
  3. $A$ is the image of an injective local diffeomorphism into $N$, i.e. there exists smooth $z$-manifold $Z$ and smooth injective immersion $H:Z \to N$ with image($H$)=$H(Z)=A$

Is this true? I'll write my attempts in an answer.

Best Answer

1 implies 2: Choose $(G,y,Y)=(\iota,n,A)$.

  • (and of course: 1 implies 3: Choose $(H,z,Z)=(\iota,n,A)$.)

3 implies 1: same reason as why 2 implies 1: Image of local diffeomorphism is open (in range).

  • (and of course: 3 implies 2: Obvious, but anyway: choose $(G,y,Y)=(H,z,Z)$.)

2 implies 3: Here, I'd combine 2 implies 1 and 1 implies 3: Choose $(H,z,Z)=(\iota,n,A)$

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