 Solution

If $M^2=Q^2$ with $M>0$, thenThe curvature singularity remains at $r=0$. At $r=M$ the static chart has a coordinate singularity, removed by the Eddington-Finkelstein coordinates above. The Kretschmann scalar there equals $8/M^4$, which is finite. The coincident horizons form a degenerate Killing horizon, and its future outer branch is the event horizon of the extremal black hole. A double zero of $F$ does not remove the horizon.