How to show that a product is a product in $\text{Set}$ Suppose $F:\text{Set}\rightarrow\text{Set}$ is a functor defined as follows:
If $X$ is a set, $F(X) = \text{the set of ultrafilters on X}$ 
And if $f:X\rightarrow Y$ is a morphism in $\text{Set}$ then 
$F(f): F(X)\rightarrow F(Y)$ by taking an ultrafilter, $\mathcal{F}$, on $X$ to the ultrafilter on $Y$ defined as $f_*\mathcal{F} = \{B\subseteq Y: f^{-1}(B)\in\mathcal{F}\}$.
I am trying to show that $$F(\pi_i): F(\Pi_{a\in I} X_a)\rightarrow F(X_i)$$
is a terminal cone (here $\pi_i$ are projections of the product defined as usual).
So for any other cone on the same family, say $f_i: C\rightarrow F(X_i)$ I need to find a unique morphism $g:C\rightarrow F(\Pi_{a\in I} X_a)$ such that $F(\pi_i)\circ g = f_i$ for each $i$.
I had an idea of defining $g(x)$ as the ultrafilter containing the filter  $\{\Pi_{a\in I} B_a : B_a \in F(X_a) \text{ for all } a\in I  \}$. But the problem here is that $g$ is not unique as we can choose from possibly multiple ultrafilters.
Question: How should I choose $g$?
 A: The result you are trying to prove is false.  If it were true, it would be saying that the functor $F$ preserves products, but this is false.  For instance, let $X$ be an infinite set and let $U$ be a nonprincipal ultrafilter on $X$.  Let $F$ be the filter on $X\times X$ generated by sets of the form $A\times A$ where $A\in U$.  Note that $F$ is not an ultrafilter: for instance, the diagonal $\Delta=\{(x,x):x\in X\}$ has nonempty intersection with every element of $F$, but is not itself an element of $F$ since it contains no rectangle with more than one element and $U$ is nonprincipal.  Now if $V$ is any ultrafilter on $X\times X$ which contains $F$, $F(\pi_1)(V)=F(\pi_2)(V)=U$.  There is more than one such $V$, since $F$ is not an ultrafilter.  So there is more than one element of $F(X\times X)$ whose image in $F(X)\times F(X)$ is $(U,U)$, and hence the maps $F(\pi_1)$ and $F(\pi_2)$ do not make $F(X\times X)$ a product of $F(X)$ and $F(X)$.
You can also see this by a counting argument.  For instance, if $I$ is countable and each $X_i$ has two elements, then $\prod X_i$ has cardinality $2^{\aleph_0}$ and so $F(\prod X_i)$ has cardinality $2^{2^{2^{\aleph_0}}}$.  On the other hand, $F(X_i)$ has just two elements for each $i$, so $\prod F(X_i)$ has only $2^{\aleph_0}$ elements.
