Let $\alpha$ be a limit ordinal. $f: \alpha \to \beta $ be strictly increase cofinal map.then $cf(\alpha) = cf(\beta)$.
refer to proof in the Kenneth Kunen's book. It's easy to show that $cf(\alpha) \leq cf(\beta)$ by composing a function from $cf(\alpha) \to \alpha$ with $f$.
To show that $cf(\beta) \leq cf(\alpha)$.(refer to the book). Let $g: cf(\beta) \to \beta$ be a cofinal map and consider a function $h:cf(\beta) \to \alpha$. define by $h(\xi) = \eta$. $\eta$ is least element such that $f(\eta) > g(\xi)$. How to show that $h:cf(\beta) \to \alpha$ is cofinal?
Here what I'm trying to proof:
Let $x \in \alpha$. We need to show that there is $\zeta \in cf(\beta)$ with $h(\zeta) \geq x$. And then I got stuck. How to do then?
HINT: For each $\xi\in\operatorname{cf}(\beta)$ we have $(f\circ h)(\xi)>g(\xi)$, so $f\circ h$ is cofinal in $\beta$. Show that if $h$ were not cofinal in $\alpha$, then $f\circ h$ could not be cofinal in $\beta$.