Let $M$ and $\overline{M}$ be Riemannian manifolds and $f : M \rightarrow \overline{M}$ an isometric immersion. Now, consider $X, Y, Z \in \mathfrak{X}(M)$ fields defined on $M$ and let $\tilde{X}, \tilde{Y}, \tilde{Z} \in \mathfrak{X}(\overline{M})$ their extensions to $\overline{M}$, defined by $$ \tilde{X} = (df) (X \circ f^{-1}).$$ I'd like to show that $$\langle \overline{\nabla}_\tilde{Y} \tilde{X},\tilde{Z} \rangle = \langle \nabla_Y X, Z \rangle,$$ where $\langle \cdot, \cdot \rangle$ denote the Riemannian metric and $\overline{\nabla}$ and $\nabla$ denote the affine connection associated to the metric on $\overline{M}$ and $M$ respectively.
I think it's a standard calculation, but I'm a bit confused with the extensions. I could write $$\langle \overline{\nabla}_\tilde{Y} \tilde{X},\tilde{Z} \rangle = \langle \nabla_Y X, \tilde{Z}^t\rangle + \langle \sigma(Y,X), \tilde{Z}^n,\rangle $$ where $\tilde{Z}^t$ denote the tangent component of $\tilde{Z}$ and $\tilde{Z}^n$ the normal component. Now, we have $$ \langle \overline{\nabla}_\tilde{Y} \tilde{X},\tilde{Z} \rangle = \langle \nabla_Y X, Z\rangle + \langle \sigma(Y,X), \tilde{Z}^n\rangle$$ and the second member must vanish, but I don't know how to prove it.
Anyone can help me? Thank you very much.