You can, no one will stop you. The problem is that the complexity is $$$O(n\alpha(n))$$$ in total. One call to path compression can take $$$\Theta(\log n)$$$ time. So suppose you have a union operation which causes that, then you have to spend $$$\Theta(\log n)$$$ time. Then you need to revert that, again in $$$\Theta(\log n)$$$ time. Then you have to do it again and so on. The complexity will be $$$O(n\log n)$$$ with or without path compression, but with path compression you do a lot of unnecessary operations, causing larger constant factor.

As I remember, most amortized complexity are broken by doing roll back operations:

• Amortization kinda like: "Oh, so the value of this function might not reduce more. Observation that the complexity is likely to be reduce each time you perform operations close to linear or something. Therefore its complexity must amortized to something smaller"

• Roll back: You have to do the work that you have done again. Or even redo and undo over and over for several times. And by going back, you can see that the average complexity on each operation may increase again.

• Say the vector<>: Each time you extend it by push_back(), emplace_back, resize or something. It will delete the old $$$2^k$$$ data and copy it to bigger $$$2^{k+1}$$$ data size. It will doing it about $$$O(2^0 + 2^1 + \dots 2^{\lfloor log_2(n) \rfloor}) = O(2^{\lfloor log_2(n) \rfloor + 1}) - 1) = O(n)$$$. But once you add such roll back operation, and say to reduce its size by smaller $$$2^{k-1}, 2^{k-2}, \dots, 2^1, 0$$$. Repeating extending and roll-backing and the complexity might be about $$$O(2^k \times n) = O(n^2)$$$ hence the amortized complexity break

Hence it this cases, it is also amortized $$$O(\alpha(n))$$$ on average but break during roll back operations. You can still achieve $$$O(\alpha(n))$$$ if you dont do any roll back operations, which is faster. However it is not likely rolling back with it is also faster.