What happens if the latitude is maximally north already? I assume there won't be any issue here?

You may be correct - if we're checking the latlng point [90, 0] (i.e. the north pole), this algorithm will I think fail to contain it in any polygon. But because we're using Cartesian polygons to begin with, there's no pole-enclosing polygon anyway. This is all moot for the polyfill usage, however, because the north pole cell at res 15 has a center latitude of 89.99999581738042, so it's not possible for this issue to arise. If we offered point-in-poly as an external function, this might be an issue, but probably one we could address in documentation.

I suppose we could address this issue with

if (lat > M_PI_2) {
  lat = M_PI_2 - DBL_EPSILON
}

but I don't think this is necessary for our current usage, and not worth the extra branch.

So the lat and lng here is provided by the caller, so this is theoretically a problem, but we just use the centerpoint of the polygon in reality:

There might be a problem with pointInsideLinkedGeoLoop, though, since it just uses the first point which could be the northernmost one:

The pointInsideLinkedGeoLoop function is looking at whether one loop is inside another. So for this issue to occur, you'd need a fairly pathological shape, where not only was one vertex of the outer loop on the north pole, but one vertex of the inner loop was also on the north pole, eg.

But again, the loops involved here are derived from H3 cell boundaries, and there is no cell with a center or a vertex on the north pole.

Perhaps add a comment explaining the assumption that only cell centers or vertexes are expected to be compared here against the user input?

Added a long comment noting the issue.

Perhaps add a comment explaining the assumption that only cell centers or vertexes are expected to be compared here against the user input?