gradient curve $\gamma$ defined on $(-T,0]$, can't be extended from $\gamma(-T)$? Let $f$ be a semi-concave function on an Alexandrov space $X$. 
Denote $\gamma_p(t)$ the $f$-gradient curve with $\gamma_p(0)=p$, i.e.
$$
\gamma^+_p(t)=\nabla_{\gamma_p(t)}f.
$$
If $X$ is a Riemannian manifold, $\nabla_{\gamma_p(t)}f=\nabla f(\gamma_p(t))$.
For any $p\in X$ and sufficiently small $t\geqslant 0$, there is $p'\in X$ such that $\gamma_{p'}(t)=p$. In other words, there exists a maximal $T>0$, $\gamma_p(t)$ can be defined on $(-T,0]$ with $\gamma_p(0)=p$. 
Let $q=\lim_{t\to -T}\alpha_p(t)$. Then $q$ may be a critical point of $f$, i.e. $\nabla_q f=0$. Or $|\nabla_q f|>0$, but if we connect gradient curve $\gamma_{q}(t)$ defined on $(-T',0]$ ($\gamma_q(0)=q$) with gradient curve $\gamma_p(t)$ at $q$, this is not a gradient curve since it's possible that
$$
\nabla_q f\neq \gamma_p^+(-T).
$$
Can one show some examples for this case?
 A: The answer depends on your definition of semi-concave functions. If you only require them to be semi-concave on geodesics then an obvious example is given by $X$ equal to the closed unit ball in $\mathbb R^n$ and $f=d(\cdot, \partial X)$. Gradient curves starting on the boundary can not be extended to the left.
If you require the function to be semi-concave on the double of $X$ then such example is impossible and gradient curves can always be extended to the left at points where $\nabla_pf\ne 0$. This follows by a standard homology argument. I'll leave you to fill the details but briefly it goes like this. Suppose we have  a point $p\in X^n$ such that $\nabla_pf\ne 0$ but the gradient curve starting at $p$ can not be extended to the left. Suppose $p\notin\partial X$. Then $H_n(X,X\backslash \{p\},\mathbb Z_2)\cong H_{n-1}(\Sigma_pX,\mathbb Z_2)\ne 0$. On the other hand since the gradient curve starting at $p$ does not extend to the left we can use the gradient flow $\phi_t$ of $f$ to push any relative cycle in $C_n(X,X\backslash \{p\},\mathbb Z_2)$ into $X\backslash \{p\}$ which implies that it's equal to zero in homology which gives a contradiction. If $p\in\partial X$ then the same argument applies to the flow of the double of $f$ on the double of $X$ which has no boundary.
