Induced Connection on $\Sigma\subset M$ Let $(M,g)$ be a Riemannian manifold, $\Sigma$ a manifold and $F:\Sigma \rightarrow M$ a smooth map. For $X,Y \in \Gamma(T\Sigma)$ vector fields and $\tilde{\nabla}$ the pull back connection on $F^*(TM)$ obtained by pulling back the Levi-Civita connection $\nabla$ on $(M,g)$, prove that
\begin{equation} \tilde{\nabla}_X F_{*}Y = \tilde{\nabla}_Y F_{*}X + F_{*}[X,Y] \end{equation}.
I think it is just a matter of unravelling definitions but I keep getting lost. Help would be greatly appreciated.
In the same vein I am having trouble proving that the pull back curvature is given by
\begin{equation} \tilde{R}(X,Y)Z = R(F_{*}X,F_{*}Y)Z \end{equation}
where $X,Y \in T_pM$ and $Z \in (F^*(TM))_p$.
 A: Define a connection : $$ \nabla_XY :=(\widetilde{\nabla}_XY)^T $$
Here $T$ means "tangent" : Let $\widetilde{\nabla}$ be a connection on $M$, and $X,\ Y$ be vector fields on $F(\Sigma)$. Then we project $  \widetilde{\nabla}_X Y (x)$ onto $T_x (F(\Sigma))$ where $x\in F(\Sigma)$.
Set $$ \widetilde{\nabla}_XY:=\nabla_XY + B(X,Y)\Rightarrow
B(X,Y)=B(Y,X)
$$ Here component of $\widetilde{\nabla}_XY$ tangent to $F(\Sigma)$ is $\nabla_X Y$ so that $B(X,Y)$ is a component of $\widetilde{\nabla}_XY$ normal to $F(\Sigma)$. In further $X,\ Y$ are tangent to $F(\Sigma)$ so that so is $[X,Y]$. Hence $B$ is symmetric.
Hence $$\begin{align*} &\widetilde{R}(X,Y)Z \\&=
 \widetilde{\nabla}_X(\widetilde{\nabla}_Y Z) -
 \widetilde{\nabla}_Y(\widetilde{\nabla}_XZ) -
 \widetilde{\nabla}_{[X,Y]} Z \\
& = \widetilde{\nabla}_X(\nabla_YZ + B(Y,Z) ) -
 \widetilde{\nabla}_Y(\nabla_X Z + B(X,Z) ) - [\nabla_{[X,Y]} Z +
 B([X,Y],Z)] \\&= \nabla_X(\nabla_YZ) + B(X,\nabla_YZ)+\widetilde{\nabla}_X B(Y,Z)
\\ &- \{ \nabla_Y(\nabla_XZ) + B(Y,\nabla_XZ)+\widetilde{\nabla}_Y
B(X,Z) - [\nabla_{[X,Y]} Z +
 B([X,Y],Z)] \\
&= R(X,Y)Z +\{  B(X,\nabla_YZ)+\widetilde{\nabla}_X B(Y,Z)
-  B(Y,\nabla_XZ)- \widetilde{\nabla}_Y B(X,Z) -
 B([X,Y],Z) \}  
\end{align*}$$
Hence $$\widetilde{R}(X,Y,Z,U)=R(X,Y,Z,U) + (\widetilde{\nabla}_X
B(Y,Z) - \widetilde{\nabla}_Y B(X,Z), U) $$
