Naturality of the pullback connection

Solution 1:

You need to show that $Xg(Y,Z)=g((\phi^\ast\tilde\nabla)_XY,Z)+g(Y,(\phi^\ast\tilde\nabla)_XZ)$.

Try showing that for $p\in M$, $$g_p(((\phi^\ast\tilde\nabla)_XY)_p,Z_p)=((\phi^{-1})^\ast g)_{\phi(p)}\left((\tilde\nabla_{\phi_\ast X}\phi_\ast Y)_{\phi(p)},(\phi_\ast Z))_{\phi(p)}\right)$$ and similarily $$g_p(Y_p,((\phi^\ast\tilde\nabla)_XZ)_p)=((\phi^{-1})^\ast g)_{\phi(p)}\left((\phi_\ast Y)_{\phi(p)},(\tilde\nabla_{\phi_\ast X}\phi_\ast Z)_{\phi(p)}\right)$$

Add these terms and use the fact that $(\phi^{-1})^\ast g=\tilde g$ and the metric compatibility of $\tilde\nabla$.

You should get that there sum at $p$ is $$ \left((\phi_\ast X)(\tilde g(\phi_\ast Y,\phi_\ast Z)\right)(\phi(p)).$$ Then try showing that this is equal to $\left(Xg(Y,Z)\right)(p)$