# Friday, September 09

:::{.remark}
Let $F: U\to \RR^k$, where $U \subseteq \RR^n$ is open (e.g. a coordinate chart for a manifold) and $n\geq k$.
Without loss of generality, assume $F(\vector 0) = \vector 0$.
Assume $F$ is smooth and suppose $DF(\vector 0): \RR^n\to \RR^k$ is onto as a linear function.
Choosing a basis for $\ker DF(\vector 0) \cong \RR^{n-k}\times\ts{0}$ yields a $k\times n$ matrix $DF(\vector 0) = [0, A]$ in block form where $A\in \GL_k(\RR)$, which up to a change of basis can be represented as $[0, I]$.
Write $x\in \RR^n \cong \RR^{n-k} \times \RR^k$ as $x = (y, z)$ and take a multivariable Taylor expansion to obtain 
\[
F(y, z) 
&= \vector 0 + DF(\vector 0)(y, z)^t + \cdots \\
&= \vector 0 + [0, I](y,z)^t + \cdots \\
&= z + \bigo(\abs{y}^2 + \abs{z}^2 )
.\]
The goal is to produce a *submanifold chart* (in Lee: a *slice chart*) for $F\inv(\vector 0)$ around $\vector 0$.
I.e. find a $U' \subseteq U$ open and $\phi: U\to \RR^n$ such that $\phi\inv(\RR^{n-k} \times \ts{0}) = \ts{x = (y,z) \in U' \st F(x) = F(y,z) = 0}$:

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Define $F: U\to \RR^n$ by $G(y, z) = (y, F(y,z))$; then $DG(y,z) = \matt I 0 {\dd{y}F}{\dd{x} F}$ and $DF(0,0) = \matt I 0 0 I$.
By the inverse function theorem (proved by the contraction mapping principle) gives an open $U' \subseteq U$ with $V' \da G(U') \subseteq \RR^n$ with a smooth $H: U'\to U$ such that $HG = \id_{U'}$ and $GH = \id_{V'}$, i.e. $G$ is a diffeomorphism onto its image in this neighborhood.
Then $\phi \da \ro{G}{U'}$ is a submanifold chart and $\phi\inv = H$, and moreover $G \circ \varphi\inv: U'\to\RR^n$ is $(y,z) \mapsto (z, F(y, z))$ and $F\circ \varphi\inv: U'\to \RR^k$ is the projection $(y,z)\mapsto z$.
Thus $F(y,z) = 0 \iff (y,z) \in \phi\inv(\RR^{n-k} \ts{0})$.
Idea: this flattens out the level set of zero along with all other level sets in a small enough neighborhood of zero in the image of $F$.
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:::{.remark}
A similar analysis works for $f: W\to \RR^n$ with $W \subseteq \RR^k$ and $k\leq n$.
Assume $W$ is a neighborhood of zero and $f(0) = 0$.
One can ask that $Df(0)\in \Mat_{n\times k}(\RR)$ is injective, so maximal rank, and by changing coordinates assume its image is $\RR^k \times \ts{0}$.
It thus takes the form $fF(0) = [I, 0]^t$ where $I$ is $k\times k$, where we've changed coordinates to replace an arbitrary invertible matrix by the identity.
Write $(y,z)\in W \times \RR^{n-k}$ and define $G: W\times \RR^{n-k} \to \RR^n$ by $g(y,z) = f(y,z) + (0, z)$.
Differentiating yields $Dg(0) = [\dd{}{y}g, \dd{}{z} g] = \matt I 0 0 I$, so again apply the inverse function theorem to obtain open sets $U' \subseteq W\times \RR^{n-k}$ and $V' = g(U') \subseteq \RR^n$ with $h:V'\to U'$ smooth with $hg = \id_{U'}, gh = \id_{V'}$.
So $h: V'\to h(V') \subseteq \RR^n$ defines a chart on the codomain of $f$ where the composition is the projection $h\circ f(y) = h\circ g(y, 0) = (y,0)$.

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:::{.remark}
To be defined more precisely later: a smooth map $f:M\to N$ between smooth manifolds is an **immersion** iff $Df(p)$ is everywhere injective on tangent spaces, and a **submersion** iff surjective.
The regular level set theorem guarantees that if $y\in N$ is a regular value of $f$, then $f\inv(\ts{y})$ is a submanifold of dimension $\dim M - \dim N$.
This does not imply that images $f(M) \subseteq N$ is a submanifold when $f$ is an immersion -- take a nodal curve.
A canonical counterexample is limiting to a node:

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