For submersion and submetry,why can we lift a geodesic "horizontally" to a geodesic? A map $\sigma:X\to Y$ between locally compact complete inner metric spaces is called a submetry if $\sigma(B_r(p))=B_r(\sigma(p))$ for all $r>0$ and $p\in X$. Why is that a geodesic in $Y$ can be lifted "horizontally" to a geodesic in $X$. And the case for submersion? The problem lies when we lift a point we don't know where we lift in the fiber.
 A: (1) Recall the definition of submersion : If $\sigma :
T_pX=V_p\oplus H_p \rightarrow T_{\sigma(p)} Y$ then $\sigma$ on
$H_p$ is isometry and $\sigma|V_p=0$ So we have $$ \sigma\circ
\overline{c}=c\Rightarrow {\rm length}\ c\leq {\rm length}\
\overline{c}\ \ast$$
(2) And if $c$ is a unique unit speed shortest geodesic in
$Y$ from $y_1$ to $y_2$, and if $\overline{c}$ is a lift with
$c=\sigma\circ \overline{c}$, then
$$l:={\rm length}\ \overline{c}={\rm length}\ c$$
If $\sigma(x_i)=y_i$ and if $\alpha$ is a shortest geodesic from
$x_1$ to $x_2$, then $${\rm length}\ \sigma\circ \alpha \leq {\rm
length}\ \alpha \leq l$$ So $\sigma\circ \alpha=c$ Hence $$ {\rm
length}\ \overline{c}={\rm length}\ c={\rm length}\ \sigma\circ
\alpha \leq {\rm length}\ \alpha$$ So $\overline{c}$
is shortest
(3) In further we have $$
 \sigma B_l(x_1) \supseteq B_l (y_1)$$ by lifting in (2) And $\ast$ implies that $ \sigma B_l(x_1) \subseteq B_l (y_1)$
(4) Assume that $ \sigma B_r(\overline{x}) =
B_r(\sigma(\overline{x})=x)$ i.e. $\sigma$ is submetry And assume
that $c$ is unit speed shortest path from $x$ to $y\in
\partial B_r(\sigma (x))$ If $r$ is small it is unique
Assume that $\overline{y}\in B_r(\overline{x}) $ s.t. $\sigma(\overline{y})=y$
If $d_X(\overline{x},\overline{y}) =R<r$, then $$
 \sigma B_R(\overline{x} )=B_R(x) $$ so that $y\in B_R(x)$ It is a contradiction
If $r$ is small so $B_\frac{r}{2} (\overline{x})\cap
 B_\frac{r}{2} (\overline{y}) $ has unique pt $\overline{z}$ Then $$
 \sigma(\overline{z})\in B_\frac{r}{2} (x) \cap
 B_\frac{r}{2} (y) $$
Hence $\sigma(\overline{z})=c(\frac{r}{2})$
Repeatedly do the same process for $c|[0,\frac{r}{2}],\
c|[\frac{r}{2},r]$ Then such $ \overline{z}$ consists of a shortest
geodesic from $\overline{x}$ to $\overline{y}$, which is a lift of
$c$
