Proving restriction is surjective

Question

Let $g:M\to M$ and $M_g = \bigcap_{n=1}^\infty g^n(M)$. Then the restriction $f = g|_{M_g}:M_g\to M_g$ is surjective if:
a)$M$ is compact and $g$ is continuous or
b)#$g^{-1}(y)<\infty$ for all $y\in M$

I have no idea how to tackle this, any tips?

Answer 1

For b)

Suppose $g^{-1}(y) \cap M_g$ is empty. Then for any $x \in g^{-1}(y)$, let $n_x$ be such that $x \notin g^{n_x}(M)$. Since $g^{-1}(y)$ is finite, then there is a maximum of all these $n_x$. Call this $n_{max}$. For this value $g^{-1}(y) \cap g^{n_{max}}(M)$ is empty. So $y \notin M_g$