Prove without coordinates that covariant derivatives are "almost" related under isometric immersion? I'm trying to solve this problem: Let $F : (M,g) \to (N,h)$ be an isometric immersion. For any $p \in M$, let $\pi_p$ be the orthogonal projection from $T_{F(p)}N$ to the image of $dF_p : T_pM \to T_{F(p)}N$. Let $X, Y$ be smooth vector fields on $M$ which are $F$-related to $\tilde X$ and $\tilde Y$ on $N$, respectively [i.e. $\forall p$, $dF_p(X(p)) = \tilde X(F(p))$]. Let $\nabla$ and $\tilde \nabla$ be the Levi-Civita connections on $M$ and $N$, repsectively. Prove that $\forall p \in M$,
$$ dF_p((\nabla_XY)(p)) = \pi_p((\tilde \nabla_{\tilde X}\tilde Y)(F(p))).$$
So, this shouldn't be hard to prove in local coordinates (and that's what I'll end up doing for my homework), but the statement seems like it ought to be provable without coordinates, but when I tried I ended up feeling like I didn't quite have enough information. More specifically, without using coordinates, all I really know about $\nabla$ is that it's symmetric and is compatible with $g$, which doesn't seem like quite enough to show something tangible about $\nabla_X Y$.
Is there a way to do this?
(Edit: it might be helpful for future searchers to know that this is very close to problem 3 of section 2.3 of do Carmo's Riemannian Geometry)
 A: Okay, here goes: Since $F$ is an immersion, $dF_p$ admits an inverse from its image, so we want to show that $(dF_p)^{-1}\circ\pi_p((\tilde\nabla_{\tilde X}\tilde Y)(F(p))$ is a connection on $M$ that is symmetric and respects $g$. That it is a connection is more or less obvious, since $(dF_p)^{-1}\circ\pi_p$ is linear at each $p$ and $\tilde\nabla$ is a connection.
So symmetry: using linearity again, we have $$(d F_p)^{-1} \circ \pi_p((\tilde\nabla_{\tilde X}\tilde Y)(F(p))) - (d F_p)^{-1}\circ\pi_p((\tilde\nabla_{\tilde Y}\tilde X)(F(p))) $$$$= (d F_p)^{-1}\circ\pi_p((\tilde\nabla_{\tilde X}\tilde Y)(F(p)) - (\tilde\nabla_{\tilde Y}\tilde X)(F(p)))$$
which is $$(dF_p)^{-1}\circ\pi_p\circ dF_p([X,Y](p)) = [X,Y](p)$$ since $[\tilde X,\tilde Y]$ is $F$-related to $[X,Y]$ and $\pi_p$ is the identity on the image of $dF_p$.
For compatibility with the metric, since $F$ is an isometric immersion, if $Z$ is $F$-related to $\tilde Z$ and $\langle,\rangle_p$ and we write $\langle\langle,\rangle\rangle_{F(p)}$ for $g(p)$ and $h(F(p))$ respectively, then we have
$$ \langle(dF_p)^{-1}\circ\pi_p(\tilde\nabla_{\tilde X}\tilde Y),Z\rangle_p + \langle Y,(dF_p)^{-1}\circ\pi_p(\tilde\nabla_{\tilde X}\tilde Z)\rangle_p$$
$$ = \langle\langle\pi_p(\tilde\nabla_{\tilde X}\tilde Y,\tilde Z\rangle\rangle_{F(p)} + \langle\langle\tilde Y,\pi_p(\tilde\nabla_{\tilde X}\tilde Z\rangle\rangle_{F(p)}$$ 
but since $\tilde Y$ and $\tilde Z$ are in the image of $dF_p$ and $\tilde\nabla$ is compatible with the metric, this is just
$$\tilde X(F(p))\langle\langle\tilde Y,\tilde Z\rangle\rangle_{F(p)} = F^\ast(\tilde X\langle\langle\tilde Y,\tilde Z\rangle\rangle)(p) = X(p)\langle Y,Z\rangle_p.$$
Since both of these hold for all $p\in M$, we've shown that our map is the Levi-Civita connection on $M$, or equivalently,
$$dF_p((\nabla_XY)(p)) = \pi_p((\tilde\nabla_{\tilde X}\tilde Y)(F(p))).$$
