I need to prove this:
Let $(M_{1}, g)$ $(M_{2}, h)$ riemannian manifolds, consider $(M_{1} \times M_{2})$ with the product metric, and $\nabla = \nabla^{1} + \nabla^{2}$. Then $\gamma=(\gamma_{1}, \gamma_{2})$ is a geodesic $\iff$ $\gamma_{1}, \gamma_{2}$ are geodesics respectively on $M_{1}, M_{2}.$
At first look, it seemed trivial to me, but then I realised it wasn't. In fact, locally $(\dot{\gamma_{1}},0),(0, \dot{\gamma_{2}}) $ are vector fields only defined along $\gamma$ and we need to extend them near $\gamma$ to calculate $\nabla_{\dot{\gamma_{i}}}\dot{\gamma_{i}},$ $i=1,2.$ My guess was extending them so that $\dot{\gamma_{1}}, \dot{\gamma_{2}}$ are constant along $M_{2},M_{1}$ respectively. In this way, we obtain using koszul formula that $\nabla_{\dot{\gamma_{1}}}\dot{\gamma_{1}},\nabla_{\dot{\gamma_{2}}}\dot{\gamma_{2}}$ have no components along $M_{2},M_{1}$ respectively. Then we conclude using $\nabla_{\dot{\gamma}}\dot{\gamma}$=$\nabla^{1}_{\dot{\gamma_{1}}}\dot{\gamma_{1}}+\nabla^{2}_{\dot{\gamma_{2}}}\dot{\gamma_{2}}$. Do you think I did it right?