I'm trying to prove the following statement.
Let $0<\overline{\alpha}\leq\alpha$ be two ordinals such that $\omega_{\overline{\alpha}}$ is the cofinality of $\omega_\alpha$. Let $f$ be a mapping of $\omega_\alpha$ into $\omega_\alpha$ such that $f(\xi)<\xi$ for any $0<\xi<\omega_\alpha$. Then there exists $\lambda_0<\omega_\alpha$ such that $f^{-1}(\lambda_0)$ has cardinality $\geq\aleph_{\overline{\alpha}}$.
I think I almost have it: From the preceding exercise, I know that $f$ cannot be divergent, i.e. I know that there exists $\lambda<\omega_\alpha$ such that $f^{-1}([0,\lambda])$ is cofinal and therefore has cardinality $\geq\aleph_{\overline{\alpha}}$. My idea was to define $\lambda_0$ to be the least ordinal $\lambda$ such that $|f^{-1}([0,\lambda])|\geq\aleph_{\overline{\alpha}}$. I would like to conclude that $|f^{-1}(\left[0,\lambda_0\right[)|<\aleph_{\overline{\alpha}}$, but I don't know how. It would be true if the cofinality of $\lambda_0$ was $<\omega_{\overline{\alpha}}$. Is there some way to show this? Am I on the right track?
I have found out that the statement is a (very) special case of a theorem called Fodor's Lemma. However its proof is quite involved and I don't have any experience with the concept "stationary", so I'm not able in a reasonable amount of time to understand it enough to reduce it to an elementary proof of the above statement. But maybe that is an easy task for an experienced set theorist.