In a Schwarzschild black hole, the singularity is spacelike. In a Kerr black hole, it is timelike.
Is there any continuous transformation between those solutions? Can the Schwarzschild solution be obtained by taking the Kerr solution in zero angular momentum limit?
Clarification: By "solution", I am mostly interested in behavior of the singularity. Is there a continuous transition from spacelike to timelike singularity?
The Kerr singularity is obtained when $ \Sigma=r^2+a^2\cos^2\theta=0 $ which means $ r=0 $ and $ \theta=\frac{\pi}{2} $.
In Cartesian coordinates, this leads to a ring in the equatorial plane with a radius equal to $ a $.
For a Schwarzschild black hole, you have $ a = 0 $ (no spin). Then the Kerr singularity becomes the center of the black hole ($ r = 0 $ with a zero radius ring) which is the Schwarzschild singularity.
Regarding your question related to the spacelike Schwarzschild singularity or the timelike Kerr singularity, I do not follow you because in the two cases the metric tensors are not defined since $ r=0 $ or $ \Sigma=0 $. In my opinion, the scalar products then can not be calculated and consequently the type of these singularities can not be determined.
Hoping to have answered your questions,
Best regards.
In Boyer-Lindquist coordinates, the Kerr metric is $$ds^2=-\frac{\Delta}{\rho^2} \left(dt-a \sin^2{\theta}~d\phi\right)^2+\frac{\sin^2{\theta}}{\rho^2}\left(\left(r^2+a^2\right)d\phi-a dt\right)^2+\frac{\rho^2}{\Delta} dr^2+ \rho^2 d\theta^2,~~~~\rightarrow(1)$$ where, $$\Delta(r)\equiv r^2-2Mr+a^2,~~~~\rightarrow(2)$$ $$\rho^2(r,\theta)\equiv r^2+a^2 \cos^2{\theta}.~~~~\rightarrow(3)$$ A curvature singularity occurs when $\rho^2=0$, which corresponds to a ring on the equatorial plane, $$r=0,~~~~\cos{\theta}=0.~~~~\rightarrow(4)$$ To judge the nature of the singularity, which is a surface of the form $\rho=\mathrm{const.}$, let us modify the coordinate system such that we have a coordinate $\rho$ instead of $r$.
Start with $$d\rho=\partial_\mu \rho~dx^\mu,~~~~\rightarrow(5)$$ where, using Eq.(3), $$\partial_\mu \rho=\left(0,\frac{r}{\rho},\frac{-a^2\cos{\theta}\sin{\theta}}{\rho},0\right).~~~~\rightarrow(6)$$ Then, substituiting Eq.(6) into Eq.(5), we get $$d\rho=\frac{1}{\rho}\left(rdr-a^2 \cos{\theta}\sin{\theta}d\theta\right).~~~~\rightarrow(7)$$ which could be written as $$\color{red}{\rho dr}=\frac{\rho}{r}\left(\rho d\rho-a^2\cos{\theta}\sin{\theta}d\theta\right).~~~~\rightarrow(8)$$ Now, substitute Eq.(8) into the original metric Eq.(1) to express the Kerr metric in coordinates $\{t,\rho,\theta,\phi\}$:
$$ds^2=-\frac{\Delta}{\rho^2} \left(dt-a \sin^2{\theta}~d\phi\right)^2+\frac{\sin^2{\theta}}{\rho^2}\left(\left(r^2+a^2\right)d\phi-a dt\right)^2+\frac{\color{red}{\frac{\rho^2}{r^2}\left(\rho d\rho-a^2\cos{\theta}\sin{\theta}d\theta\right)^2}}{\Delta}+ \rho^2 d\theta^2,~~~~\rightarrow(9)$$
where now $$r=r(\rho,\theta)=\sqrt{\rho^2-a^2\cos^2{\theta}},~~~~\rightarrow(10)$$ and $$\Delta(r)=\Delta\left(r\left(\rho,\theta\right)\right)=\rho^2-a^2\cos^2{\theta}-2M\sqrt{\rho^2-a^2\cos^2{\theta}}+a^2.~~~~\rightarrow(11)$$ Since we only care about the nature of the $\rho$ coordinate as we approach the singularity, let us set $t=t_0,~\theta=\pi/2,~\phi=\phi_0.$ The metric then becomes $$ds^2=\frac{\frac{\rho^2}{r^2}\left(\rho d\rho\right)^2}{\Delta}=\left(\frac{\rho^2}{\rho^2-2M\rho+a^2}\right) d\rho^2.~~~~\rightarrow(12)$$ As you approach the singularity, $\rho$ gets closer and closer to zero, and $a^2$ dominates in the denominator: $$ds^2\approx\left(\frac{\rho^2}{a^2}\right) d\rho^2>0 \Longrightarrow \boxed{\lim_{\rho\rightarrow0}{ds^2\big|_\left\{t=t_0,~\theta=\pi/2,~\phi=\phi_0\right\}}>0.}~~~~\rightarrow(13)$$ This shows that the coordinate $\rho$ is a spatial coordinate all the way down to the singularity.
If, however, we let $a\rightarrow0$ to get the Schwarzschild limit, then we recover the usual behavior of Schwarzschild: $\rho$ (which is identical to $r$ when $a=0$) becomes timelike below the horizon and all the way down to the singularity. We have $$ds^2=\left(\frac{\rho^2}{\rho^2-2M\rho}\right) d\rho^2=\left(\frac{\rho}{\rho-2M}\right) d\rho^2\Longrightarrow\boxed{\lim_{\rho\rightarrow0}{ds^2\big|_\left\{\color{red}{a=0},~t=t_0,~\theta=\pi/2,~\phi=\phi_0\right\}}<0.}~~~~\rightarrow(14)$$
Equation (14) is the limit of equation (13) as $a\rightarrow0$ in the manner you asked for.
