# Why $\dim(\ker T_z f)=\dim(T_z(f^{-1}(c)))$?

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 $$n-p$$: $$$$\dim T_z \left( f^{-1}(c) \right) = n-p$$$$
Next, the formula in the rank theorem shows you that the derivative of $$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$$ to conclude that
$$$$\dim \left( \ker (T_z f) \right) = n-p$$$$