= Solution
{c}
Put $d\Omega_2^2=d\theta^2+\sin^2\theta\,d\varphi^2$, the standard <metric tensor> on the <unit sphere>. From $t=v+m/\lambda$ one has $dt=dv-m\lambda^{-2}d\lambda$. The apparently divergent terms cancel:
$$
\boxed{g=-\lambda^2dv^2+2m\,dv\,d\lambda+m^2d\Omega_2^2.}
$$
The $(v,\lambda)$ <metric tensor> block has <determinant> $-m^2\ne0$, independently of $\lambda$. Its coefficients are <real analytic>, so this formula defines a regular <Lorentzian metric> on
$$
\boxed{M_1=\mathbb R_v\times\mathbb R_\lambda\times S^2.}
$$
The polar-coordinate degeneracies at the poles are handled by ordinary <two-sphere> charts. Thus $\lambda=0$ is a <coordinate singularity> of the original chart and is crossed by this <spacetime extension>.
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