This is motivated by Exercise 15(c) in Cori, Lascar, Pelletier.
Given a map $f:\beta\to \alpha$ whose image is not strictly bounded (i.e. the upper bounded of the image of $f$ is the ordinal $\alpha$). Let $\delta$ be the least ordinal in the set $\{\gamma\le \beta:\sup_{\xi\in \gamma} f(\xi) = \alpha\}$.
Define a map $g:\delta\to \alpha$ as: $\gamma\mapsto \sup_{\xi \in \gamma}f(\xi)$
The solution manual then says the image of $g$ has no strict upper bound. I am intuitively convinced, but failed writing out a detailed proof. Any help, please?
Since $\delta$ is the least one such that $\sup f``\delta=\alpha$, we have that $g$ is a well-defined function.
Next, note that $g(\gamma)\geq f(\gamma)$ for all $\gamma<\delta$, that is an easy induction to write. And therefore, since $f\restriction\delta$ already is unbounded below $\alpha$, $g$ must be too.