tangent space as kernel of map

Question

In the following notes: http://www.math.toronto.edu/mgualt/courses/MAT1300F-2016/docs/1300-2016-notes-5.pdf

during the proof of proposition 3.2, why is the $ker(Df(p))=T_pK$ ?

Answer 1

The vectors in $T_p K$ are the vectors at $p$ which are tangent to $K$. In the proof, $K$ is assumed to be (locally) a level set of the function $f$, meaning the function $f$ is constant along $K$. If you take a vector $v$ based at $p$, it represents a "directional derivative" operator on functions. Specifically, for a function $g$, you have $v \, g = Dg \cdot v$. If that vector happens to be in $T_p K$, it means it is a directional derivative in a direction tangent to $K$. Again, $f$ is constant along $K$, so the directional derivatives in these directions (tangent to $K$) should intuitively be zero when applied to $f$. This means $v \in T_p K$ should be in the kernel of $Df$.

Answer 2

First the relevant part of the setup from Theorem 3.2:

$M$ is a $n$-dimensional smooth manifold and $K$ is a regular sub-manifold of co-dimension $k$. The sub-manifold $K$ is the level-set of a regular function $f:M\rightarrow N$ with constant rank mapping in some other smooth manifold $N$.

An answer to your question fitting into the linked lecture (see Definition 2.11 and Proposition 2.12):

There are charts $(\varphi,U),(\psi,V)$ with $U\subset M$ and $V\subset N$ in which $\varphi(U)$ is an open subset $\Omega$ of $\mathbb{R}^n$, $\varphi(K\cap U)$ is the intersection of $\Omega$ with the subspace spanned by the first $n-k$ vectors of the canonical basis, i.e., $\mathbb{R}^{(n-k)}\times\{0\}^k$, and $f$ is essentially the local projection onto the last $k$ components of its mapped argument, i.e., $\psi\circ f \circ \varphi^{-1}: (x_1,\ldots,x_n)\mapsto (0,\ldots,0,x_{n-k+1},\ldots,x_n)$.

In this coordinates the spaces $T_p K$ and $\ker(D f(p))$ are both trivially equal to the subspace $\mathbb{R}^{(n-k)}\times \{0\}^k$.