Smooth Manifolds - Key Proof

We present the proof of the inverse function theorem for manifolds, demonstrating how the classical result from multivariable calculus lifts to the manifold setting through charts.

ProofInverse Function Theorem for Manifolds

Statement: Let $f: M \to N$ be a smooth map between $n$ -dimensional manifolds. If $df_p: T_pM \to T_{f(p)}N$ is an isomorphism, then there exist neighborhoods $U$ of $p$ and $V$ of $f(p)$ such that $f|_U: U \to V$ is a diffeomorphism.

Proof: Choose charts $(U_1, \varphi)$ centered at $p$ and $(V_1, \psi)$ centered at $f(p)$ such that $\varphi(p) = 0$ and $\psi(f(p)) = 0$ . Consider the coordinate representation

F = \psi \circ f \circ \varphi^{-1}: \varphi(U_1 \cap f^{-1}(V_1)) \to \psi(V_1)

This is a smooth map between open subsets of $\mathbb{R}^n$ . The chain rule gives:

dF_0 = d(\psi)_{f(p)} \circ df_p \circ d(\varphi^{-1})_0

Since $\varphi$ and $\psi$ are diffeomorphisms (charts), their differentials are isomorphisms. By hypothesis, $df_p$ is also an isomorphism. Therefore $dF_0$ is an isomorphism.

By the classical inverse function theorem in $\mathbb{R}^n$ , there exist open neighborhoods $W$ of $0$ in $\mathbb{R}^n$ and $W'$ of $0$ such that $F|_W: W \to W'$ is a diffeomorphism.

Now set:

$U = \varphi^{-1}(W) \subset M$
$V = \psi^{-1}(W') \subset N$

Then $f|_U = \psi^{-1} \circ F|_W \circ \varphi$ , which is the composition of three diffeomorphisms:

$\varphi: U \to W$ (restriction of chart)
$F|_W: W \to W'$ (inverse function theorem in $\mathbb{R}^n$ )
$\psi^{-1}: W' \to V$ (inverse of chart)

Therefore $f|_U: U \to V$ is a diffeomorphism. $\square$

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Remark

The proof illustrates the general strategy in differential geometry: reduce problems to the Euclidean case using charts, apply classical results from calculus, then pull back to the manifold. The smoothness of transition functions ensures this strategy is coherent.

ExampleComputing the Differential

For $f: S^2 \to S^2$ given by $f(x,y,z) = (x,y,-z)$ (reflection through the equator), we can compute $df_p$ using stereographic coordinates. At the north pole $N = (0,0,1)$ , the differential acts as $df_N(v) = (v_1, v_2, -v_3)$ in standard coordinates, which is clearly an isomorphism.

ProofImplicit Function Theorem - Sketch

When $f: M^m \to N^n$ is a submersion at $p$ (meaning $df_p$ is surjective), we can choose adapted coordinates such that $f$ looks locally like a projection. Specifically:

Choose coordinates $(x^1, \ldots, x^m)$ near $p$ and $(y^1, \ldots, y^n)$ near $q = f(p)$
By the constant rank theorem, after a change of coordinates, $f$ has the form

f(x^1, \ldots, x^n, x^{n+1}, \ldots, x^m) = (x^1, \ldots, x^n)

The level set $f^{-1}(q)$ is then locally given by $x^1 = \cdots = x^n = 0$ , which is clearly a smooth $(m-n)$ -dimensional submanifold with coordinates $(x^{n+1}, \ldots, x^m)$ . $\square$

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The implicit function theorem converts analytic conditions (surjectivity of the differential) into geometric conclusions (smoothness of level sets). This is one of the most frequently used techniques for constructing manifolds.