Isometric map of geodesic Assume a Riemann manifold $(M,g)$ and a smooth map $\sigma:M\times M\rightarrow M$, $(m_{1},m_{2})\rightarrow \sigma_{m_{1}}(m_{2})$, such that:
$\forall m\in M$  $\sigma_{m}:M\rightarrow M$ is an isometry and $\sigma_{m}=m$;
$\sigma_{m}\circ \sigma_{m}=1_{M}$;
$D\sigma_{m}|_{T_{m}M}=-1_{T_{m}M}$;
This is part of a definition of a symmetric space. But other part is irrelevant for the problem I have. Now, fix $m\in M$, and let $\gamma (t)$ be a geodesic of $M$ such that $\gamma (0)=m$. Show that:
$$\sigma_{m}\gamma (t)=\gamma (-t).$$
I proceeded as follows:
$$D(\sigma_{m}\circ \gamma (t))=D\sigma_{m}\circ\gamma^{'}(t).$$ 
Since $D\sigma_{m}|_{T_{m}M}=-1_{T_{m}M}$, we have:
$$D(\sigma_{m}\circ \gamma (t))=-\gamma^{'} (t).$$
Since $-\gamma^{'} (t)=D\gamma (-t)$, we have:
$$D(\sigma_{m}\circ \gamma (t))=D\gamma (-t).$$
So, I assume, that we have this equality:
 $$\sigma_{m}\circ \gamma (t)=\gamma (-t).$$
But in the question there was an assumption that $\gamma (t)$ is geodesic and I didn't use it. So my question is: is my line of thinking is correct, or I just completely wrong?
 A: Your mistake is the following: You have $D(\sigma_m \circ \gamma)(t) = D\sigma_m \gamma'(t)$. Also $\gamma'(t) \in T_{\gamma(t)}M$ but on the other hand $D\sigma_m = -Id$ holds only on $T_mM = T_{\gamma(0)}M$. Thus you can only deduce $$D(\sigma_m \circ \gamma)(0) = -\gamma'(0).$$ For other values of $t$ the equation $D(\sigma_m \circ \gamma)(t) = -\gamma'(t)$ makes no sense, since $D(\sigma_m \circ \gamma)(t) \in T_{\sigma_m(\gamma(t))}M$ and $-\gamma'(t) \in T_{\gamma(t)}M$. For the same reason $T_{\gamma(t)}M \ni -\gamma'(t) = D\gamma(-t) \in T_{\gamma(-t)}M$ makes no sense either.
The correct argument goes as follows: Since $\sigma_m$ is an isometry it follows that $\sigma_m \circ \gamma$ is also a geodesic. Moreover 
$$D(\sigma_m \circ \gamma)(0) = D\sigma_m\gamma'(0) = -\gamma'(0) = d/dt_{t = 0}\gamma(-t).$$
Thus $t \mapsto \gamma(-t)$ and $t \mapsto \sigma_m(\gamma(t))$ are both geodesics with the same initial conditions. Hence they coincide (here we use that we are dealing with geodesics). 
