The claim below is considered obvious in Levy's Basic Set Theory:
If $\kappa$ is a singular cardinal, then there is a regressive function $f:\kappa\to \text{cf}(\kappa)$, such that for all $\gamma<\text{cf}(\kappa)$, the preimage $f^{-1}"\{\gamma\}$ is bounded below $\kappa$.
I don't see why this claim holds. The intuitive thing to try first is to take a cofinal sequence $(\alpha_i\mid i<\text{cf}(\kappa))$ in $\kappa$ with ordertype $\text{cf}(\kappa)$, and think of this sequence as breaking $\kappa$ into intervals. So we would map each $x$ in the interval $(\alpha_i, \alpha_{i+1}]$ to $\alpha_i$. But this doesn't work, because it is unclear where $\alpha_\lambda$ themselves will be mapped to, for any limit ordinal $\lambda$. Enumerating the limit points in $(\alpha_i\mid i<\text{cf}(\kappa))$ and trying a similar map doesn't seem to help either, because we'll need to worry about limits of limits, and so on.
(Edit: as you can see, I was completely confusing myself above. I was trying to map things to the elements on the cofinal sequence, instead of their indices. As a result the map I was trying to define has range way above $\text{cf}(\kappa)$!)
So I am a little bit lost now. Do we need to use Fodor's lemma somehow?
The regressive condition is trivial to satisfy since $\operatorname{cf}(\kappa)<\kappa$: just let $f(\alpha)=0$ for all $\alpha\leq\operatorname{cf}(\kappa)$. Then, you can pick a sequence $(\alpha_i)_{i<\operatorname{cf}(\kappa)}$ with limit $\kappa$ and define $f(\alpha)$ to be the least $i$ such that $\alpha<\alpha_i$ if $\alpha>\operatorname{cf}(\kappa)$.