In a solution to a question of an exam I found the following claim:
Let $(M,g)$ a Riemannian manifold and let $\nabla$ be the Levi-Civita connnection induced by the metric $g$. Consider $(\overline{M},\overline{g})$ a submanifold of $M$ of codimension $1$, where $\overline{g}$ is the induced metric by $g$. Let $\overline{\nabla}$ be the Levi-Civita connection for $\overline{g}$. Let $\chi(\overline{M})$ be the space of smooth vector fields on $\overline{M}$. We can define the second fundamental form $k$ as a $(0,2)$ tensor by defining
$k(X,Y):= -g(\nabla_{X}n,Y) \qquad X,Y \in \chi(\overline{M})$
where $n$ is the vector normal to $\overline{M}$.
Until here everything is fine. The claim is the following: For $X,Y,Z \in \chi(\overline{M})$
$\nabla_{X}k(Y,Z)= k(\nabla_{X}Y,Z)+ k(Y,\nabla_XZ)= k(\overline{\nabla}_{X}Y,Z)+ k(Y,\overline{\nabla}_XZ)= \overline{\nabla}_{X}k(Y,Z)$
I Cannot understand the claim. Indeed, it seems to me that they are using a reasoning as follows:
$\nabla_XY= \overline{\nabla}_X Y$
whenever $X,Y \in \chi(\overline{M})$. But as far as I know this is not necessarily true. The claim needs to be true because it is part of the solution, but I can not understand why. Can anyone help me?