Schwarzschild: Proof that $\{r<2m\}$ is a black hole I saw the following proof to show that $\{r<2m\}$ is a black hole in the Schwarzschild metric. 
Consider the Schwarzschild metric:
$$
g=-V(r)\text d t^2 + \frac{1}{V(r)}\text d r^2 + r^2 \text d \Omega^2\;,\quad V(r)=1-\frac{2m}{r}\;.
$$
Introduce the Eddington-Finkelstein coordinate $v=t+f(r)$ where $f'=1/V$. Thus the metric reads
$$
g=-V(r)\text d v^2 + 2\text d v \text d r + r^2 \text d \Omega^2\;.
$$
Now consider a causal curve $\gamma(s)=(v(s),r(s),\theta(s),\varphi(s))$ such that $g(\dot \gamma,\dot \gamma)\leq 0$. We then have
$$
g(\dot \gamma,\dot \gamma)\leq 0 \quad \Leftrightarrow \quad -V(r)\dot v^2 + 2 \dot v \dot r \leq  0 \quad \Leftrightarrow \quad \dot v (-V(r)\dot v + 2 \dot r ) \leq  0\;.
$$
Now they claim that $\dot v$ is positive since the standard choice of time orientation in the exterior region corresponds to $\dot v > 0$. From this we can directly conclude $\dot r \leq  0$ for $r<2m$ since $V(r)<0$. Thus $\{r<2m\}$ is a black hole.
But I do not understand the argument that provides $\dot v > 0$. By definition we have $\dot v = \dot t + \frac{\dot r}{V(r)}$. I do not see why this is always positive. Can someone help me?
 A: Choosing $\dot{v}>0$ is equivalent to choosing an arrow of time in your spacetime. Purely from GR, there is now way to determine the arrow of time and therefore it is not possible to prove $\dot{v}>0$. The logic is the following: You observe that there are two equivalence classes of timelike vector fields (which you call future- and past-directed). You have to make sure that the splitting of future- and past-directed timelike vector fields is a coordinate invariant statement! In your physical description you require objects to follow future-directed timelike curves (or null curves for massless objects). You cannot show this but it is an assumption that leads you to a well-defined causal structure (see here).
And clearly, past-directed geodesics can of course escape the black hole. They are just assumed to be unphysical.
For the case of time-coordinates $v$ and $t$, the following diagram shows you that the definition of future directed does not change inside the lightcone upon changing the coordinate: In the lightcone, $v>0$ and $t>0$ coincide.

A: As you mentioned, $\gamma$ is a causal curve.
The space-time causal structure is given by the semi-definite (pseudo-Riemannian) metric. But locally we can always choose a Galilean basis in which $g_{\mu \nu}$ is just the Minkowski tensor (this is called the equivalence principle, as you probably know).
The structure of the Minkowski space arises naturally on the tangent space: we have the splitting of the tangent space into three zones: future light cone, past light cone and the one outside the light cone (space-like intervals).
Now the question becomes: how should one describe these zones mathematically in an arbitrary basis (with arbitrary semi-definite $g_{\mu \nu}$)?
This is simple: the tangent space vectors lying outside of the light cone are defined via the relation
$$ g_{\mu \nu} A^{\mu} A^{\nu} < 0 $$
(or $> 0$ if you choose the $- + + +$ signature). What we are left with are the future-pointing and past-pointing tangent space vectors. How to separate these two kinds? There aren't really a good way to do this since we don't have an arrow of time yet.
So let's introduce one. We can do this by simply choosing the future (like $A^0 > 0$) and the past (accordingly, $A^0 < 0$). Note that this definition should coincide between different space-time points.
In your question, the time coordinate is modified a bit to give $v$. Note from the structure of your metric that $A^v > 0$, along with $A_{\mu} A^{\mu} > 0$ defines one of the internal regions of the light cone. Everything else is as I described. I hope this helped.
