I am studying submanifolds and I have some problems with the proof of a claim about Rank Theorem.

Rank Theorem: Let $f: U \subseteq \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a smooth map and each point has the constant rank $0 < p \leq \min\{m,n\}$. Then, for every point $x\in U$, there is a chart $(V,\varphi)$ in $\mathbb{R}^n$ around $x$, and there is a chart $(W,\psi)$ in $\mathbb{R}^m$ around $f(x)$ that: $$\begin{aligned}\psi \circ f \circ \varphi^{-1}:\varphi(V)&\rightarrow \psi(W) \\ (z_1, \cdots,z_n) &\mapsto (z_1,\cdots, z_p,o_{\mathbb{R}^{m-p}}) \end{aligned}$$

Now with the assumptions of Rank Theorem, we have two claims:

  1. For every $c \in f(U)$, $f^{-1}(c)$ is a $(n-p)$-dimensional submanifold of $\mathbb{R}^n$.
  2. For every point in a level surface of $f$, like $z=(z_1,\cdots,z_n)\in f^{-1}(c)$: $T_z(f^{-1}(c))=\ker(T_z f)$

The proof of the first claim is obvious, but I have a problem with the second one.

Proof (claim 2): Using the first claim, we know that $f^{-1}(c)$ is a $(n-p)$-dimensional submanifold of $\mathbb{R}^n$. Now we investigate the tangent space of $f^{-1}(c)$ in a point like $z \in f^{-1}(c)$.

Let $\gamma: I \rightarrow f^{-1}(c)$ be a smooth curve in $f^{-1}(c)$ that passes from $\gamma(0)=z$. Then the vector $[\gamma]_z=T_z(f^{-1}(c))$ is a tangent vector of this submanifold in $z$.

Since $\gamma(I)\subseteq f^{-1}(c)$, so $f\circ \gamma :I \rightarrow \mathbb{R}^m$ that takes every $t\in I$ to $c$, is the constant map $f \circ \gamma \equiv c \in \mathbb{R}^m$. So: $$\frac{d}{dt}(f\circ \gamma) |_{t=0} =0 \Rightarrow T_0 (f \circ \gamma)=T_{\gamma(0)} f \circ T_0 \gamma \equiv 0$$ Thus: $$T_0 \gamma:T_0 I \rightarrow T_z(f^{-1}(c)) \Rightarrow T_z f(T_0 \gamma(0,1))=0 \Rightarrow [\gamma]_z \in \ker T_z f$$ So, $T_z(f^{-1}(c)) \subseteq \ker T_z f$.

But since $\dim(\ker T_z f)=\dim(T_z(f^{-1}(c)))$ and both are $(n-p)$-dimensional, $T_z(f^{-1}(c)) = \ker T_z f$ and the claim is proved.

Actually, I have a problem with the last line of this proof. I don't find the reason of the equality of dimensions of these two spaces.

Any help is appreciated.


$f^{-1}(c)$ is an $(n-p)$ dimensional submanifold so the dimension of its tangent space at a point $z$ is also $n-p$: \begin{equation} \dim T_z \left( f^{-1}(c) \right) = n-p \end{equation}

Next, the formula in the rank theorem shows you that the derivative of $f$ at a point $z$, $Df(z)$ has rank $p$; hence $T_zf$ (which is simply $Df(z)$ but thought of as a map between tangent spaces) also has rank $p$. Now apply the rank-nullity theorem to $T_zf$ to conclude that

\begin{equation} \dim \left( \ker (T_z f) \right) = n-p \end{equation}


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