This is a follow-up to my earlier question on the Kunen exercise, which asks for a proof that any two well-orderings on an infinite cardinal $\kappa$ agree on a set of size $\kappa$. I'm perfectly willing to accept proofs that use choice and I understand how to use transfinite recursion to get the result for regular cardinals.
The proof I've seen that I come closest to understanding is the answer given by Jason Palumbo here. He notes that the problem is equivalent to showing that if $f:\kappa \to \kappa$ is bijective, then there is a set $A \subseteq \kappa$ on which $f$ is increasing.
If $\kappa$ is singular, it is the limit of (successor) cardinals $\lambda_\alpha$ with cofinality greater than $\operatorname{cf} \kappa$.
Let $f: \kappa \to \kappa$ be a bijection. Then $\exists A_1 \subseteq \lambda_1$ with $f$ increasing on $A_1$.. Define $\gamma_1 = \sup f(A_1)$.
Here's the step I don't understand. Jason states that $\operatorname{cf} \gamma_1 = \operatorname{cf} \lambda_1 \gt \operatorname{cf} \kappa$. I don't see why the first equality has to hold. If it is the case, it follows that $\gamma_1 \lt \kappa$ and we can complete the proof by recursion.
So if someone would be kind enough to explain this for me, I'd very much appreciate it. Thanks.