Differential geometry-geodesics Let $\nabla_1$ and $\nabla_2$ be two affine connections of manifolds $M_1$ and $M_2$ and $\nabla$ induced connection on product $M_1\times M_2$. Prove the following:
If $\gamma_1$, $\gamma_2$ are curves on $M_1$, $M_2$ and $X_1$, $X_2$ are vector fields parallel along $\gamma_1$, $\gamma_2$, then the field $d_{j_1}(X_1)+d_{j_2}(X_2)$ is parallel along $\gamma_1 +\gamma_2$, and conversely also holds (every field parallel along $\gamma_1 +\gamma_2$ has this form). 
Can we conclude that geodesics on $M_1\times M_2$ are products of geodesics on $M_1$ and $M_2$?
Detailed solutions and explanations are welcome. Thanks in advance.
 A: For $M=M_1\times M_2$ we have $TM=TM_1 \times TM_2$; at each point $p=(p_1,p_2)\in M$ the tangent space $T_pM$ naturally splits as the direct sum $T_{p_1}M_1\oplus T_{p_2}M_2$.  
Now, if $\nabla^i$ are connections on $M_i, i=1,2$, then the standard connection on $M$ is the sum of two connections $\nabla^1 + \nabla^2$ meaning that for vector fields $X_i, Y_i, i=1,2$ on $M$ and $X=(X_1,X_2), Y=(Y_1,Y_2)$, we have
$$
\nabla_X(Y)= \nabla^1_{X_1}(Y_1) + \nabla^2_{X_2}Y_2. 
$$ 
Suppose that $c=c(t)=(c_1(t),c_2(t))$ is a smooth curve in $M$ with $c_i$ smooth curves in $M_i, i=1,2$. Then (in view of the formula for $\nabla$ above)
$$
\nabla_{c'(t)} c'(t)= \nabla^1_{c_1'(t)} c_1'(t) + \nabla^2_{c_2'(t)}c_2'(t). 
$$ 
The curve $c$ is a geodesic iff $\nabla_{c'(t)} c'(t)=0$. Note that a vector
$v=(v_1, v_2)\in T_{(p_1,p_2)}(M)$ is zero if and only if both components of $v$ are zero. Therefore,
$$
\nabla_{c'(t)} c'(t)=0 \iff \nabla^1_{c_1'(t)} c_1'(t) =0,  \nabla^2_{c_2'(t)}c_2'(t)=0. 
$$
Hence, $c$ is a geodesic if and only if both curves $c_1, c_2$ are geodesics.  
