Let $X$ be a set, let $\mathscr{F}$ be the collection of filters on $X$, and let $\mathscr U$ be the collection of ultrafilters on $X$. Define $\pi: \mathscr F \to 2^{\mathscr U}$ by $$\pi(F) = \{U \in \mathscr U: U \supset F\}.$$ By the ultrafilter lemma, $\pi (F)$ is nonempty.
Is $\pi$ a bijection? In other words, can every filter be identified with its set of ultrafilter extensions?
It seems like the answer should be no just based on cardinality considerations, but I'm not sure how to show it. In case the answer is no: What is the right way to think about the relationship between a filter and the set of its ultrafilter extensions?
First of all note that $\pi$ is injective. Let $F_{1},F_{2}$ be two different filters. Then wlog there is a set $C\in F_{1}$ such that $C\not\in F_{2}$. Then there is an ultrafilter $U$ which is an extension of $F_{2}$ containing $C^{c}$. (A basic result on the extension of filters.)
The problem comes with surjectivity. Let $X$ be infinite and consider the set $$\mathcal{A}=\{U\subset 2^{\mathcal{U}}:U\text{ principal}\}.$$ Recall that $U$ is principal if there exists a set $A\subset X$ such that $B\in U$ if and only if $A\subset B\subset X$. It is well known that principal Ultrafilters are all of the form $$U_{x}=\{A\subset X:x\in A\}$$ with $x\in X$ and that for a infinite set $X$ there exist non-principal ultrafilters, which are extensions of the Frechet filter.
Now suppose there is a filter $F$ such that $\pi(F)=\mathcal{A}$. If $A\in F$ with $A\neq X$ then let $y\in A^{c}$ and note that $X\setminus\{y\}\in F$, which implies $U_{y}\not\in\pi(F)$. So $F=\{X\}$ is the trivial filter, for which clearly $\pi(F)=\mathcal{U}$.
In conclusion, every filter has a unique set of ultrafilter extensions, but not every set of ultrafilters is uniquely defined by a filter.