I am correct.
CanonFanBoy is correct. Although a f/2 lens is an f/2 lens, a M43 sensor is smaller than a FF sensor and hence M43 f/2 lens is slower than a FF f/2 lens. You have to apply the multiplication factor to get the 35mm equivalent.
A slower lens is a lens that lets in less light! A M43 lens with the f-stop as a FF lets in less light than the FF lens.
The f-stop is simply the focal length divided by the aperture diameter. For example, 100mm focal length divided by pupil dimeter of 50mm = f/2. Divided by a pupil diameter of 25mm, it would be f/4. This is literally the mathematical definition of the f-number (N = f/D).
The brightness of the projected scene has absolutely nothing to do with the size of the sensor. If you use a sensor smaller than the projected image circle, you'll just cut off (crop) the image. However, it would be accurate to say that using a smaller sensor behind the same optics will capture fewer photons (of course).
Instead of comparing sensors of the same aspect ratio, imagine if you took a FF sensor, and cut it in vertically in half down the middle. The left half sensor would now capture exactly half the light (number of photons) as the whole sensor. But the image and exposure settings on the left half sensor would be exactly the same as that of the whole sensor, and the image on the left half sensor would be the same as the image on the whole sensor, if you just cropped off the left half.
From a practical perspective, if you're using a Sekonic light meter to measure your exposure settings given your strobe setup... you don't dial in whether you are using a MFT or FF sensor. Given the amount of light you're throwing on your subject, the exposure settings will be the same regardless of MFT/APSC/FF.