Version of Fodor's Lemma I know that pressing down functions exist, but it is not intuitively clear to me. Is there a way to prove the existence of such functions in this form:
$$f: \omega_1 \to \omega_1 $$ $$\exists \alpha \in \omega_1. f^{\to}(\alpha) \subseteq \alpha$$ (image of $\alpha$)
Also, would that be useful to prove this tricky version of pressing down lemma:
If  $f: \omega_1 \to \omega_1$ and $f(\alpha) < \alpha$ then $\exists \beta \in \omega_1. |f^{\gets}\{\beta\}=\omega_1|$
(pre-image of $\beta$)
Is there a straightforward way of showing $\exists \beta \in \omega_1. |f^{\gets}\{\beta\}=\omega_1|$ without using clubs?
 A: The existence of such $f$ is trivial: take the identity function, then $f^{-1}(\alpha)=\alpha$, for any $\alpha$. 
So this is indeed not useful for proving that any regressive function has a large fiber. But note that Fodor's lemma states more than just cardinality. It tells you that there is a "large fiber", where large here means stationary.
To prove Fodor's lemma, you need to essentially remember the definition of a diagonal intersection, and the theorem that the diagonal intersection of $\omega_1$ clubs (in $\omega_1$) is a club again. Using this the proof more or less writes itself: suppose $f$ is a regressive function which is a counterexample, choose clubs witnessing this, take the diagonal intersection, it is non-empty, and show that any $\alpha$ in that non-empty set implies a contradiction.
(Morally speaking, this is the right proof as well, since without the axiom of choice Fodor's lemma holds for $\omega_1$ if and only if the diagonal intersection of any family of $\omega_1$ sets which contain a club also contains a club.)

Here is an argument showing that the cardinality of a fiber has to be $\omega_1$, it does not appeal directly to Fodor's lemma, although it hides the use of clubs.
Suppose that every fiber is countable, then for every $\alpha<\omega_1$, there is some $\beta$ such that if $\gamma>\beta$, then $f(\gamma)>\alpha$.
Pick some $\alpha_0$, and let $\alpha_{n+1}$ be the least ordinal such that $f(\beta)>\alpha_n$ for all $\beta>\alpha_{n+1}$. Now consider $\alpha=\sup\{\alpha_n\mid n<\omega\}$. Then by the fact $f$ is regressive, $f(\alpha)<\alpha$. On the other hand, if $f(\alpha)<\alpha$, then there is some $n$ such that $f(\alpha)<\alpha_n$. This is now a contradiction, since $\alpha>\alpha_{n+1}$, so $f(\alpha)>\alpha_n$.
