Assume that function $f:\omega_1\to\omega_1\times\omega_1$ is bijection. Prove that set $C=\{\alpha<\omega_1:f|\alpha:\alpha\to\alpha\times\alpha$ is bijection$\}$ is closed and unbounded.
This question was asked already (Prove that set $C=\{\alpha<\omega_1:f|\alpha:\alpha\xrightarrow{\rm 1:1,onto}\alpha\times\alpha\}$ is closed and unbounded.) but I have in mind another solution using following fact: for any function $g:\kappa^n\to\kappa$ set: $$ M_g=\{\alpha<\kappa:g[\alpha^n]\subseteq\alpha\} $$ is closed and unbounded.
I have doubts if I know exactly how to use it. Do I need just to take $g=f^{-1}\circ f$?
Since $f$ is injective, it is only a matter of finding the points where it is also surjective.
Now, suppose that $C$ is not a club, then there is some $S$, a stationary subset of $M_f$, disjoint from $C$.
Define the following function $g(\alpha)=\min\{\beta<\alpha\mid\beta\notin f[\alpha^2]\}$. Then $g$ is well-defined on $S$ and it is regressive there. By Fodor's lemma, $g$ is constant on a stationary subset $S'$ with value $\eta$. But this means that for no $\alpha\in S'$ there is some $(\beta,\gamma)\in\alpha^2$ such that $f(\beta,\gamma)=\eta$. This is a contradiction to the fact that $f$ is itself a bijection.