Show that $[\widetilde{X},\widetilde{Y}]^H=\widetilde{[X,Y]}$ and that $[\widetilde{X},W]$ is vertical if $W$ is vertical. This is problem 9 from chapter 5 in Riemannian Manifolds: An introduction to Curvature.
Suppose $p:(\widetilde M,\widetilde g)\rightarrow (M,g)$ is a Riemannian submersion. 
We denote $\widetilde X$ for the horizontal lift of $X$, i.e, the unique horizontal $\widetilde X$ such that $p_*\widetilde X_x=X_{p(x)}$.
a) Show that $[\widetilde{X},\widetilde{Y}]^H=\widetilde{[X,Y]}$ and that $[\widetilde{X},W]$ is vertical if $W$ is vertical.
b) Let $\widetilde\nabla$ and $\nabla$ be riemannian connections in $\widetilde M$ and $M$. Show that
\begin{equation}
(1) \ \ \ \widetilde\nabla_\widetilde x\widetilde Y=\widetilde{\nabla_X Y}+\frac{1}{2} [\widetilde X,\widetilde Y]^V
\end{equation}
For a) I think it would be useful if $p_*[\widetilde X,\widetilde Y]=[p_*\widetilde X,p_*\widetilde Y]$ was valid. But as far as I know, this is valid only in the case that $p$ is a diffeomorphism. The fact that we need to take the horizontal part of $[\widetilde X,\widetilde Y]$ is confusing to me as well. How to see that the commutator of two horizontal fields is not always horizontal?
For b), my goal is to calculate $\langle\widetilde\nabla_\widetilde X\widetilde Y,\widetilde Z\rangle$, where $\widetilde Z$ is any horizontal field. I do the same for a vertical field $W$. By taking these products and using that $\widetilde g(\widetilde X, \widetilde Y)=g(p_*X,p_*Y)$, we want to show that they are equal to taking the product with the horizontal and vertical parts of $(1)$.
Taking the product with the horizontal part is almost the same as doing $\langle \widetilde{\nabla_XY},\widetilde Z\rangle$, but for them to be equal I need to have $\widetilde X\langle \widetilde Y,\widetilde Z\rangle=X\langle Y,Z\rangle$. Is this correct? If so, how to show that this is equal?
Taking the product with a vertical vector field resulted in $\langle\widetilde\nabla_\widetilde X\widetilde Y,W\rangle = \frac{1}{2}(\langle W, [\widetilde X,\widetilde Y]^V \rangle - W\langle \widetilde X, \widetilde Y\rangle)$, but I'm not sure what to do with this second term. Did I make a mistake somewhere?
 A: Ok, I think I got it after some massive help from my friend Hugo. 
a) First, let's show that $p_*[\widetilde{X},\widetilde{Y}]=[X,Y](f)$ for all $f\in C^\infty (M)$:
\begin{align*}
p_*[\widetilde{X},\widetilde{Y}](f)& =[\widetilde{X},\widetilde{Y}](f\circ p) \\
& = \widetilde{X}(\widetilde{Y}(f\circ p))-\widetilde{Y}(\widetilde{X}(f\circ p)) \\
& = \widetilde{X}(p_*\widetilde{Y}(f))-\widetilde{Y}(p_*\widetilde{X}(f)) \\
& = \widetilde{X}(Y_{p(x)}(f))-\widetilde{Y}(X_{p(x)}(f)) \\
& = \widetilde{X}(Y(f)\circ p)-\widetilde{Y}(X(f)\circ p) \\
& = p_*(\widetilde{X})(Y(f))-p_*(\widetilde{Y})(X(f)) \\
& = XY(f)-YX(f) \\
& = [X,Y](f) \\
\end{align*}
Now we note that $[\widetilde{X},\widetilde{Y}]=[\widetilde{X},\widetilde{Y}]^H+[\widetilde{X},\widetilde{Y}]^V$ and that $p_*[\widetilde{X},\widetilde{Y}]^V=0$. Thus, 
\begin{align*}
    p_*[\widetilde{X},\widetilde{Y}]^H_{x}=[X,Y]_{p(x)}
\end{align*}
concluding the proof.
Now to show that $[\widetilde{X},W]$ is vertical if $W$ is vertical. By hypothesis, $p_*(W)=0$ and $p_*(\widetilde{X})=X$, so we have:
\begin{align*}
    p_*[\widetilde{X},W](f)&=[\widetilde{X},W](f\circ p) \\
    & =\widetilde{X}(W(f\circ p))-W(\widetilde{X}(f\circ p)) \\
    & =\widetilde{X}(p_*W(f))-W(p_*\widetilde{X}(f)) \\
    & = 0 - W(X(f)\circ p) \\
    & = -p_*W(X(f)) \\
    & = 0
\end{align*}
concluding the exercise.
b) By the Kozsul formula, where $\widetilde{Z}$ is a horizontal vector:
\begin{align*}
    \langle (\widetilde\nabla_{\widetilde{X}} \widetilde{Y})^H,\widetilde{Z}\rangle & = \frac{1}{2}(\widetilde{X}\langle \widetilde{Y},\widetilde{Z}\rangle+\widetilde{Y}\langle \widetilde{X},\widetilde{Z}\rangle-\widetilde{Z}\langle \widetilde{X},\widetilde{Y}\rangle + \\
    & + \langle[\widetilde{X},\widetilde{Y}],\widetilde{Z}\rangle-\langle[\widetilde{X},\widetilde{Z}],\widetilde{Y}\rangle-\langle[\widetilde{Y},\widetilde{Z}],\widetilde{X}\rangle) \\
\end{align*}
Now, we show that $X\langle Y,Z\rangle =\widetilde{X}\langle \widetilde{Y},\widetilde{Z}\rangle$ (where I was having problem). This is indeed the case:
\begin{align*}
    X\langle Y,Z\rangle&=p_*\widetilde{X}\langle Y,Z\rangle \\
    & = \widetilde{X}(\langle Y,Z\rangle \circ p) \\
    & = \widetilde{X}\langle \widetilde{Y},\widetilde{Z}\rangle
\end{align*}
By using this together with the fact that $\langle[\widetilde{X},\widetilde{Y}],\widetilde{Z}\rangle=\langle[\widetilde{X},\widetilde{Y}]^H,\widetilde{Z}\rangle=\langle [X,Y],Z\rangle$ (as shown in the previous exercise) and $p$ being a Riemannian submersion. Substituting in the Kozsul, we have
\begin{align*}
    \langle (\widetilde\nabla_{\widetilde X} \widetilde Y)^H, \widetilde{Z}\rangle & = \frac{1}{2}(X\langle Y, Z\rangle+Y\langle X, Z\rangle- Z\langle X,Y\rangle + \\
    & + \langle[X,Y], Z\rangle-\langle[X, Z],Y\rangle-\langle[Y, Z],X\rangle) \\
    & = \langle \nabla_{X} Y, Z\rangle = \langle \widetilde{\nabla_{X} Y}, \widetilde{Z}\rangle
\end{align*}
where we used once again that $p$ is a Riemannian submersion. Since this is valid for all $\widetilde(Z)$, we have shown that $(\widetilde\nabla_{\widetilde X} \widetilde Y)^H=\widetilde{\nabla_{X} Y}$.
Now for the vertical part. By first noting that $\langle\widetilde\nabla_{\widetilde{X}} \widetilde{Y},W\rangle = \langle (\widetilde\nabla_{\widetilde{X}} \widetilde{Y})^V,W\rangle$, we calculate 
\begin{align*}
    \langle (\widetilde\nabla_{\widetilde{X}} \widetilde{Y})^V, W\rangle & = \frac{1}{2}(\widetilde{X}\langle \widetilde{Y}, W\rangle+\widetilde{Y}\langle \widetilde{X}, W\rangle- W\langle \widetilde{X},\widetilde{Y}\rangle + \\
    & + \langle[\widetilde{X},\widetilde{Y}], W\rangle-\langle[\widetilde{X}, W],\widetilde{Y}\rangle-\langle[\widetilde{Y}, W],\widetilde{X}\rangle) \\
\end{align*}
and by using that $W$ is vertical, $[\widetilde{X}, W]$ e $[\widetilde{Y}, W]$ are vertical (by the previous exercise). Our expression results in
\begin{align*}
    \langle (\widetilde\nabla_{\widetilde{X}} \widetilde{Y})^V, W\rangle & = \frac{1}{2}(- W\langle \widetilde{X},\widetilde{Y}\rangle + \langle[\widetilde{X},\widetilde{Y}]^V, W\rangle \\
\end{align*}
So now we better show that $W\langle \widetilde{X},\widetilde{Y}\rangle=0$. First note that $\langle \widetilde{X},\widetilde{Y}\rangle$ is constante in the fiber. In fact, let $q_1,q_2\in \widetilde{M}_{p(x)}=p^{-1}x$, $x\in M$, then by the Riemannian submersion
\begin{align*}
    \langle \widetilde{X},\widetilde{Y}\rangle (q_1)=\langle X,Y\rangle (p(x)) = \langle \widetilde{X},\widetilde{Y}\rangle (q_2)
\end{align*}
As $W$ is vertical, it follows that $W\langle \widetilde{X},\widetilde{Y}\rangle=0$. Thus, we have
\begin{align*}
    \langle (\widetilde\nabla_{\widetilde{X}} \widetilde{Y})^V, W\rangle & = \frac{1}{2}(\langle[\widetilde{X},\widetilde{Y}]^V, W\rangle \\
\end{align*}
proving that $(\widetilde\nabla_{\widetilde{X}} \widetilde{Y})^V=\frac{1}{2}[\widetilde{X},\widetilde{Y}]^V$, concluding the exercise.
