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How is the preimage theorem proved for abstract manifolds (not necessarily embedded in $\mathbb{R}^N$)?

It states that if $F:M\rightarrow N$ is a smooth map between smooth manifolds and $q\in N$ and $Z=F^{-1}(q)\neq\emptyset$ and $F_*|_p$ (i.e. $D_pF$, or $dF_p$, or various other notations!) is surjective for all $p\in F^{-1}(q)$, then $Z$ is an embedded submanifold of $M$ of dimension $k=\mathrm{dim}(M)-\mathrm{dim}(N)$.

I've got as far as showing that for fixed $p\in F^{-1}(q)$, the kernel $K$ of $F_*|_p$ in $T_pM$ must be a subspace of dimension $k$. If $M\subset\mathbb{R}^N$, we can then proceed by taking a linear map $\mu:\mathbb{R}^N\rightarrow\mathbb{R}^k$ such that $\mathrm{ker}(\mu)\cap K=\{0\}$ and then considering the map $(F,\mu)$. But what if we can't assume this? Must there be a map $G:M\rightarrow\mathbb{R}^k$ such that $G_*|_p$ satisfies the condition required for $\mu$ above, or is there a different method?

Many thanks for any help with this!

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I think you are relying too much on the fact that M is embedded. However, there is a much easier way to go around this. (I will only sketch the proof, so you can learn yourself by filling out the details) First, I will assume that you have seen proof of Implicit Function Theorem for $\mathbb{R}^n$ to $\mathbb{R}^m$. You can use this fact to prove preimage theorem for $\mathbb{R}^n$ to $\mathbb{R}^m$. Now, consider arbitrary manifolds, say $M$ and $N$. Then, take local chart (so say $\psi \circ f \circ \varphi^{-1}$), apply preimage theorem for $\mathbb{R}^n$ to $\mathbb{R}^m$. Then, you have a submanifold $K$ of $\varphi(U) \subset \mathbb{R}^n$ where $U$ is the domain of local chart for $M$. Last step is to show that $\varphi^{-1}(K)$ is a submanifold of $M$ with required dimension. Since showing a set is a manifold is to check local conditions (other than tedious second countability and hausdorff), this finishes off the proof.

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To add, this proof that I sketched only works for $\mathbb{R}$ manifolds, probably not on $\mathbb{C}$ manifolds, although I believe that the idea is the same. –  chhan92 Jan 10 '13 at 21:56
    
The statement and your proof are valid for complex manifolds: just replace $\mathbb R$ by $\mathbb C$ everywhere! (The key point is that the implicit mapping theorem is valid for holomorphic=complex differentiable maps) –  Georges Elencwajg Jan 10 '13 at 22:01
    
Is it? Ok. But how about proof of Implicit Function Theorem for $\mathbb{C}$ manifolds? I think it is different because of holomorphic charts? –  chhan92 Jan 10 '13 at 22:03
1  
You can find the easy proof of the complex implicit function theorem in Grauert-Fritzsche's book here, page 33, a page which Google graciously lets you consult. –  Georges Elencwajg Jan 10 '13 at 22:33
    
wow it is basically the same proof, watching the fact that maps are complex differentiable. Thank you –  chhan92 Jan 10 '13 at 22:55

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