tangent space of manifold and Kernel I want to understand the proof of this Theorem :
Theorem: 
Let $ S $ be a submanifold of $ E $ and let $ a \in S $. Assume that there exists a submersion $f : E \supset U  \rightarrow F$  of class $ \textit{C}\ ^{1} $ and $ b\in f(U) $ such that $ S=f^{-1}(b) $ then $ T_{a}S=Ker f'(a) $.
Proof: 
Since $ f(S)\subset \{b\}$, $f'(a).T_{a}S\subset \{0\}$. Hence $T_{a}S\subset Ker f'(a)$. Since both spaces have the same dimension, the result follows.
My problem is that I am not able to understand all the steps in the proof. Can some one help me?
 A: In differential geometry, there is the following "well known"
Theorem 1: Let $f:E\supset U \rightarrow F$ be a smooth submersion (just as in your question). Let $b \in f(U)$. Then the set $S:= f^{-1}(\{b\})$ is in fact a smooth submanifold of $E$ with $\dim S = \dim E - \dim F.$
The proof of Theorem 1 considers representations of $f$ in local coordinates around points $a \in S$ and invokes the Implicit Function Theorem which in turn is "well known" from multivariate calculus (or see here: http://en.wikipedia.org/wiki/Implicit_function_theorem).
Recall that $f$ being a submersion means $rank\ f' = \dim F$ (or equivalently $f'$ is surjective) at every point $b \in U$. This property of $f$ is essential for the application of the Implicit Fuction Theorem.
You should find Theorem 1 in any good textbook on differential geometry or manifold theory, and the Implicit Function Theorem in any good textbook on multivariate calculus.
It seems to me that the theorem in your question assumes Theorem 1, since it states that $S$ is a submanifold.
Now that we have $\dim S,$ it's not so difficult anymore to prove the claim about the tangent spaces $T_aS$ for $a\in S.$ So let's fix an $a\in S$ and pick a tangent vector $v\in T_aS.$ We want to show that $f'(a)v = 0.$ One way to do this is as follows. Since $v$ is a tangent vector at $S$ in $a$ there are an open interval $J \subseteq \mathbb R$ such that $0 \in J$, and a smooth curve $\gamma:J\rightarrow S$ such that $\gamma(0) = a$ and $\gamma'(0) = v.$ Since $f(S) = \{b\}$ and $\gamma(J) \subseteq S,$ we get $f\circ\gamma(J) = \{b\},$ or in other words, $f\circ \gamma$ is constant. Since the differential of a constant map is $0$, we get by using the chain rule
$$
0 = (f\circ \gamma)'(0) = f'(\gamma(0))\gamma'(0) = f'(a)v,
$$
just as we wanted to show. Since $v \in T_aS$ was arbitrary, we now have
$$
f'(a)T_aS = 0\quad or\ equivalently \quad T_aS \subseteq ker\ f'(a).
$$
Now we know from Theorem 1 that $\dim S = \dim E - \dim F$. And since $f$ is a submersion, we have $rank\ f'(a) = \dim F$. This implies by some linear algebra that
$$
\dim ker\ f'(a) = \dim E - rank\ f'(a) = \dim E - \dim F = \dim S = \dim T_aS.
$$
So we have that "both spaces have the same dimension", as stated in the proof of your question. Again by linear algebra, from the fact that $T_aS \subseteq ker\ f'(a)$ together with the equality of dimensions, we conclude
$$
T_aS = ker\ f'(a),
$$
as desired.
