Thanks for the explanation. Could you post what the equation(s) is? I'd be interested in looking at it.
What Alan is describing is the ability to crop beyond the crop factor of your format.
The common formats with their dimensions and following area and crop factor play a role here:
FF
36 mm * 24 mm = 864 mm^2
Diagonal of 43.27 mm
Crop factor 1 (Reference)
Canon APS-C
22.2 mm * 14.8 mm = 328.56 mm^2
Diagonal of 26.68 mm
Crop factor 1.62 (43.27/26.68)
Mirco Four Thirds
17.3 mm * 13.0 mm = 224.9 mm^2
Diagonal of 21.64 mm
Crop factor 2 (43.27/21.64)
The different areas mean that for a given lens, each sensor format will show a different field of view, regardless of resolution. But on the larger formats, given enough resolution you can crop down so that you are left with an area that's the size of one of the smaller formats.
The term resolution is ambiguous. It can mean the absolute number of pixel in an image, for instance. Or it can be a measure of actual detail if it is given as a density.
Cropping an APS-C image down to a MFT format reduces the total resolution down to 224.9 / 328.56 = 0.685 ~ 68.5 %. Meaning if your total APS-C resolution is 32.5 MP (EOS M6 II, EOS 90D) you'd end up with 22.3 MP. Which is more than MFT has to offer and why more megapixels on an identical sensor size can deliver more detail should be intuitive.
An alternative way to get to this right away is to simply look at the pixel density or like Alan did, the size of each pixel. For a 32.5 MP Canon APS-C that would be 328,56 * 10^-6 m / (32.5 * 10^6) = 10.109 Micrometer^2, and taking the square root it get's you 3.18 Micrometer which is the width and height of a pixel that Alan gave. There's some rounding error involved.
So square root of (sensor area divided by number of pixels).