Well-orderded set $P$ cannot be order-isomorphic to the set of predessors of $p \in P$

Question

My lecture notes on ZFC set theory prove the following theorem

and then note that

However, I feel like the proof is wrong. What I don't understand is why we can apply the Lemma in the proof of the Corollary. More precisely, the order-isomorphism in that proof is not from the set $P$ to itself.

First I thought that we only need to restrict $f$ to $pred(p)$, but $pred(p)$ does not contain $p$, so that does not help.

Could anybody help me out, please? Am I missing anything obvious here or is the proof flawed?

Answer 1

To some extent, this is basically a longer version of what was alredy discussed in the comments.

If you have a function from $P$ to some subset of $P$, you can simply change the codomain to the larger set $P$ and you get a function again. If you want to be very precise, you probably should denote with a different symbol - but for brevity using the same symbol might have advantages (at least, unless in the proof you need to distinguish between the two functions).

In any case, if you take $f\colon P \to \operatorname{pred} p$ and the extension $\widehat f\colon P\to P$ defined by $$\widehat f(x)=f(x)$$ then you can apply the lemma to the function $\widehat f$.

These posts are related to restricting or enlarging codomain of a function:

• Is there a name for when we extend the codomain of a function? (sort of like the opposite of the restriction) on Mathematics.
• What's the notation for a function restricted to a subset of the codomain? on MathOverflow