Rudin-Keisler order 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 $a<b$ implies $f(a)\leq f(b)$.
 A: What does $g$ look like?  It's constant on some finite block $0, \dots , n_0$, then it's some larger constant value on another finite block $n_0+1, \dots , n_1$, etc.  It would be nice if there's some $A \in \mathcal{U}$ such that $A$ meets each of those blocks in at most one point, because $g$ would be injective if restricted to such an $A$.  (This might remind you of one of the characterizations of Ramsey ultrafilters - I think being minimal amongst free ultrafilters in the Rudin-Keisler ordering is equivalent to being Ramsey if I remember correctly).  Now consider the function which strictly decreases on each block such that all the values it takes on the first block are less than all the values it takes on the second block, etc.  So the first few values it takes are: 
$$n_0, n_0-1, \dots, 0, n_1, n_1 - 1, \dots, n_0 +1, \dots$$
This function has to be $\mathcal{U}$-equivalent to a weakly increasing function.  The only way this can happen is if there's an $A$ like the one we desire!
