Showing a functor from $\text{Set}$ to $\text{Set}$ is full and faithful. Suppose $F:\text{Set}\rightarrow\text{Set}$ is a functor defined as follows:


*

*If $X$ is a set:
$F(X) = \text{the set of ultrafilters on }X$ 

*If $f:X\rightarrow Y$ is a morphism: 
$F(f): F(X)\rightarrow F(Y)$ by taking an ultrafilter, $\mathcal{F}$, on $X$ to the ultrafilter on $Y$ defined as $f_*\mathcal{F} = \{B\subseteq Y: f^{-1}(B)\in\mathcal{F}\}$.
Is this functor full and faithful?
 A: It's very very far from being full.  A simple way to see this is just a counting argument.  If $X$ is any infinite set, say of cardinality $\kappa$, then $|F(X)|=2^{2^{\kappa}}$.  There are $|F(X)|^{|F(X)|}=2^{2^{2^\kappa}}$ different functions $F(X)\to F(X)$ and only $|X|^{|X|}=2^{\kappa}$ functions $X\to X$.  Since $2^{2^{2^\kappa}}>2^\kappa$, not every function $F(X)\to F(X)$ can come from a function $X\to X$.
(In fact, all this counting argument needs is that there is a single set $X$ such that $|F(X)|\geq 2^{|X|}$, since then you get $|F(X)|^{|F(X)|}\geq 2^{2^{|X|}}>2^{|X|}=|X|^{|X|}$.  You can easily see this is true for $X=\mathbb{Q}$, for instance, since for each $r\in\mathbb{R}$ there is an ultrafilter on $\mathbb{Q}$ which converges to $r$, and these ultrafilters must be distinct for distinct $r$.)
Here's another way to see it.  Note that if $X$ has one point, then $F(X)$ has one point (the unique principal ultrafilter).  Given a point $y\in Y$, the map $f:X\to Y$ whose image is $y$ then gives the map $F(f):F(X)\to F(Y)$ whose image is the principal ultrafilter corresponding to $Y$.  If $F$ were full, that would mean that every map $F(X)\to F(Y)$ has this form: that is, every element of $F(Y)$ is a principal ultrafilter.
