For the Schwarzschild solution $ds^2=-(1-\frac{2GM}{r})dt^2+(1-\frac{2GM}{r})^{-1}dr^2+r^2(d\theta^2+sin^2\theta d\phi^2)$, if we set $dr^2_*=(1-\frac{2GM}{r})^{-2}dr^2$(in other words, $r_*=r+2GMlog{|\frac{r-2GM}{2GM}|}$), then for light rays travelling in the radial direction (which means the solid angle part is 0), we have $\frac{dr_*}{dt}=\pm1$. We see that null, radial geodesics are given by $t\pm r_*=const.$ We introduce a pair of null coordinates $$ v=t+r_*\qquad and \qquad u=t-r_*\quad. \qquad (1) $$ We can see that $$ v=const.\qquad and \qquad u=const.\qquad (2) $$ represent ingoing and outgoing light rays respectively.
Replace $t$ with $t=v-r_*(r)$ in Schwarzschild metric, we have
$$
ds^2=-(1-\frac{2GM}{r})dv^2+2dvdr+r^2d\Omega^2_2.\qquad (3)
$$
The condition $u=const.$ now can be written as
$$
v=2r_*+const.=2r+4GMlog{|\frac{r-2GM}{2GM}|}+const.\qquad (4)
$$
Since $r_*$ sits in the range $r_*\in(-\infty,+\infty)$ outside the horizon, and in the range $r_*\in(-\infty,0)$ inside the horizon, we may not use $r_*$ and $t$ to dipict the graph of $v=const.$ and $u=const.$. Instead, we define
$$
v=t+r_*=t_*+r \qquad (5)
$$
and dipict the graph of $v$ and $u$ by $t_*$ and $r. It looks like this:
where the ingoing null geodesics $v=const.$ are shown in red, outgoing $u=const.$ in blue. We know that the light cone is define by the area bounded by ingoing and outgoing, future-pointing null geodesics, so we can predict that if some particle enters the horizen, it can never come out again since in the light cone inside the horizon, $r$ does not cannot increase.
Above are from Section 6.1.3 in David Tong's note. My question is, why can we directly replace $t$ with $t_*$? I admit that when defining time with $t_*$, we can get some reasonable results very naturally. However, I think that at least we should show that $t_*$ is an time like coordinate first. I used (5) to calculate the derivative $\frac{dt}{dt_*}$, and outside the horizon, the derivative is positive, but in the horizon, $$ \frac{dt}{dt_*}=\frac{-r}{2GM-r},\qquad (6) $$ which is negative. This means that inside the horizon, when we go along the $t_*$ axis in the positive direction, we are in fact going in the negative direction of the $t$. Then we shouldn't get the above conclusions if I am right. The light cones inside the horizon should also point downward.
Also, if we take (5) into (3), we can find that $t_*$ inside the horizon, just like $t$, is still not a timelike coordinate, due to the coefficient $-(1-\frac{2GM}{r})$. Then why can we still consider it as time even if it is not well defined?
I know these two questions are in contradiction with each other a little bit. I appreciate very much if someone can explain this.