Vtyjkyuk

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 $ker(Df(p))=T_pK$ ?

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 $Df$.

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 $f:Mrightarrow N$ 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 $Usubset M$ and $Vsubset N$ in which $varphi(U)$ is an open subset $Omega$ of $mathbbR^n$, $varphi(Kcap U)$ is the intersection of $Omega$ with the subspace spanned by the first $n-k$ vectors of the canonical basis, i.e., $mathbbR^(n-k)times0^k$, and $f$ is essentially the local projection onto the last $k$ components of its mapped argument, i.e., $psicirc f circ varphi^-1: (x_1,ldots,x_n)mapsto (0,ldots,0,x_n-k+1,ldots,x_n)$.

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