If $\gamma _1$ and $\gamma _2$ are two different geodesic from $x$ to $y$, then, there is no minimal geodesic from $x$ to $z$ Let $(M,g)$ a Riemanian manifold and $x,y\in M$. Suppose there is two different geodesic $\gamma _1,\gamma _2$ that connect $x$ and $y$. Show that no one of these two geodesic are minimizing after $y$. **The solution goes like :** By contradiction, suppose $\gamma _1,\gamma _2$ are two geodesic from $x$ to $y$ and suppose that $\gamma _1$ can be prolonge in a geodesic that minimal from $y$ \to $z$. Let call the portion $\gamma _2$. Then we construct a minimal geodesic from $x$ to $z$ by following first $\gamma _2$ and then $\gamma _3$. But this curve can't be smooth where as a geodesic is smooth, what is a contradiction. **Question :** Why this curve can't be smooth ?

If the tangent vectors of $\gamma_1$ and $\gamma_2$ are equal at a common point then they cannot be distinct geodesics (that point and vector serve as initial data to a system of ODEs, which has a unique solution). Therefore their derivatives are different at $y$.

If we assume $\gamma_1 \cup \gamma_3$ is smooth, the tangent vector along this path has zero covariant derivative, and at $y$ it is not equal to the tangent of $\gamma_2$ at $y$: $$ \gamma_2'(y) \
eq \gamma_1'(y) = \gamma_3'(y). $$

Apparently the curve $\gamma_2 \cup \gamma_3$ is not differentiable at $y$.