Q:

A characterization of tangent space to level set of a smooth submersion

This is corollary 5.39 in John Lee's Introduction to Smooth Manifolds.

Corollary 5.39 Suppose $S \subseteq M$ is a level set of a smooth submersion
  $\Phi=(\Phi^1,...\Phi^k): M \to \mathbb{R^k}$. A vector $v \in T_pM$
  is tangent to $S$ if and only if $v\Phi^1=...=v\Phi^k=0$.

He said that the proof is immediate. But I can't figure it out.The author said it's a restatement of Proposition 5.38 in a special case in which the defining function takes its value in $\mathbb{R^k}$.

Proposition 5.38 Suppose $M$ is a smooth manifold and $S \subseteq M$ s an embedded
  submanifold. If $\Phi:U \to N$ is any local defining function map for
  S, then $T_pS = \mathrm{Ker}d\Phi_p:T_pM \to T_{\Phi(p)}N$ for each $p
 \in S\cap U$.

The step making me stuck is the sufficiency part. 
Suppose that $v\Phi^1=...=v\Phi^k=0$, in order to use Proposition 5.38, I want to show that $v \in \mathrm{Ker}d\Phi_p$, that is, $\forall f \in C^{\infty}(\mathbb{R}^k)$, $v(f\circ \Phi)=0$. But I don't know how to reach this...The author said this proof is immediate, so may I miss something? Thanks for your help in advance.

A:

Well, you have that ${\rm d}\Phi_p(v) = ({\rm d}\Phi^1_p(v),\ldots, {\rm d}\Phi^k_p(v)) = (v\Phi^1,\ldots, v\Phi^k)$. So $v \in \ker {\rm d}\Phi_p$ if and only if $v\Phi^1 = \cdots = v\Phi^k = 0$.

