User blog comment:Deedlit11/Ordinal Notations IV: Up to a weakly inaccessible cardinal/@comment-11227630-20170301040811/@comment-5529393-20170302125319

Hmm, yes, that does make a difference. But, I get that $$\psi_\Omega(\phi(1,1,0)+1) = \phi(1,1,0)$$, and so in general $$\psi\Omega(\alpha) = \phi(1,1,0)$$ for all $$\phi(1,1,0) \le \alpha \le \Omega$$.

It would be interesting to add $$\alpha \in C(\alpha,\beta)$$ back in, too get back to 1-1 functions.