The Kerr metric is
\begin{equation} ds^2 = - \big(1-\frac{2GMr}{\rho^2}\big)dt^2-\frac{2GMra}{\rho^2}\sin^2 \theta \big(dtd\phi+d\phi dt\big)+ \frac{\rho^2}{\Delta}dr^2+\rho^2 d\theta^2 + \frac{\sin^2 \theta}{\rho^2}\big[(r^2+a^2)^2-a^2\Delta\sin^2\theta\big]d\phi^2 \end{equation}
where
\begin{equation} \rho^2= r^2+a^2\sin^2\theta \\ \Delta = r^2 -2GMr + a^2 \end{equation}
The interesting positions are the points where $g_{rr}\rightarrow \infty$ and the ones where $g_{tt}\rightarrow 0$ since they are related with the surfaces where certain Killing vectors change from spacelike to timelike or viceversa (I know that making this kind of statement about the metric is actually coordinate-dependent but all books, like Carroll, or Misner, do this to find the horizons). The interesting radii are
\begin{equation} g_{rr}\rightarrow \infty \ \ \ \ \ \text{at} \ \ \ \ \ R^{(r)}_{\pm}=GM \pm \sqrt{(GM)^2-a^2}\\ g_{tt}\rightarrow 0 \ \ \ \ \ \text{at} \ \ \ \ \ R^{(t)}_{\pm}=GM \pm \sqrt{(GM)^2-a^2\cos^2\theta} \end{equation}
If I understand correctly, the biggest one is $R^{(t)}_{+}$, which is the beginning of the ergosphere, also called the stationary limit surface or the infinite redshift surface. Then we get $R^{(r)}_{+}$ which is the outer horizon, where the escape velocity becomes greater than c. Then we have the inner horizon $R^{(r)}_{-}$ where the metric goes "back to normal" in the sense that the radial component is spacelike again so you can exit the black hole.
This is the usual treatment in all the books I checked (Carroll, Wald, Misner, etc.). But no one seems to talk about $R^{(t)}_{-}$. What's the deal with that radius? Does it have some special property? Is it just an artifact of the coordinates we are using? Is it meaningless in some other way because it's too deep into the Black Hole?