I have to prove this: Fixed $\mathcal{U}$ a non-principal ultrafilter then (1) imples (2) where:
1) Every $f:\mathbb{N}\rightarrow\mathbb{N}$ is $\mathcal{U}$-eq. to a weakly increasing function.
2) $\mathcal{U}$ is minimal in the Rudin-Keisler order.
Here what I have done:
Suppose to have an ultrafilter $\mathcal{V}$ such that $\mathcal{V}\leq\mathcal{U}$, then by definition there exists $f:\mathbb{N}\rightarrow\mathbb{N}$ such that $\mathcal{V}=f_*(\mathcal{U})$. We know that $f\cong_\mathcal{U} g$ where $g$ is a weakly increasing function. Then $f_*(\mathcal{U})=g_*(\mathcal{U})$ and so $\mathcal{V}=g_*(\mathcal{U})$. If I prove that there exists $A\in\mathcal{U}$ such that $g|_A$ is injective, then $g_*(\mathcal{U})\cong\mathcal{U}$ and so $\mathcal{V}\cong\mathcal{U}$.
So is it true that if $g$ is a weakly increasing function $\mathbb{N}\rightarrow\mathbb{N}$ and $\mathcal{U}$ a non-principal ultrafilter, then there exists $A\in\mathcal{U}$ such that $g|_A$ is injective?
EDIT: with "weakly increasing" I mean $a implies $f(a)\leq f(b)$.