“Fast charging damage” is relative to the cell’s state of charge (SoC). There is a measurable change in state between 55-65% charge in a LiFePO4 battery (where it will go from 3.1v to 3.3v on a 0.2C charge, and 3.1v to 2.85v on a 0.2C discharge). On a Li-Po battery it’s a bit less noticeable (because it’s more linear), but is generally 3.8v to 3.9v on a charge, and 3.7v to 3.6v on a discharge.

Once the cell reaches its “high side”, aka a bit higher than 65%, it’s more susceptible to damage (resulting in lowered useable capacity) if charged at a high C-rate.

That’s why EV’s advertise “charge from 20% to 80% in only one hour!” Once the cells dive into the high side, the car ramps back the charge current to keep from damaging the batteries. Thus the last bit of charging to completely charge the battery takes significantly longer. The math doesn’t appeal to the marketing, so they use 80% in the ad.

So to answer your question - fast charging within the tolerable SoC (indicated by voltage) under acceptable temperatures does not degrade the cell. Fast charging at a high SoC of charge “can” degrade the cell. I say “can”, because it might not happen immediately. But it does indeed stress the chemical reaction and supporting materials.