Suppose that $V$ is a model of $\sf ZFC$, and fix some regular $\kappa$, say $\omega\_1$ for practical purposes.


Let $\cal U$ be an ultrafilter on $\omega\_1$ in $V$ which is non-principal and even uniform. Let $\Bbb P$ be a forcing such that:


1. $\Bbb P$ does not add bounded subsets to $\kappa$, but its generic defines a canonical unbounded subset of $\kappa$, say $G$.
2. $\Bbb P$ satisfies $\kappa^+$-c.c.
3. $\Bbb P$ is homogeneous.


Of course that $\cal U$ is not an ultrafilter in $V[G]$ since it is no longer closed under supersets. But we can ask whether or not $\mathcal U\cup\{G\}$ is an ultrafilter base.



> 
> Is there a condition on $\cal U$ (or an additional condition on $\Bbb P$ or $G$) which guarantee that $\mathcal U\cup\{G\}$ is an ultrafilter base in $V[G]$?
> 
> 
> 



When can you canonically extend an ultrafilter after forcing?
Here is a necessary and sufficient condition:


**Theorem.** If $U$ is an ultrafilter in $V$, then the following
are equivalent:


1. $U\cup\{G\}$ is an ultrafilter base.
2. $G$ is not disjoint from any element of $U$ and $U$ decides every subset of $G$ in $V[G]$, meaning that for every $A\subset G$ in $V[G]$
there is $B\in U$ with $B\cap G$ contained in $A$ or disjoint from $A$.


The proof is straightforward.


It follows that many of the usual forcing notions do not have this
property. For example, if the forcing is adding a Cohen subset
$G\subset\kappa$, then we definitely won't have condition 2. For example, if $A$ consists of every other element of $G$ and $B$ is any unbounded ground model set, then it is dense that $B\cap G$ is neither contained in nor disjoint from $A$.


