Powerset functor weakly preserves pullbacks. Could you please suggest, how to prove that the covariant powerset functor $Set \rightarrow Set$ weakly preserves pullbacks? I don't get how to show weakness:
suppose we have a pullback square
$$
\begin{array}{ccc}
D & \rightarrow & B \\
\downarrow & & \downarrow \\
A & \rightarrow & C
\end{array}
$$
where $D$ is a pullback (i.e. it is unique up to isomorphism). It also can be viewed as a terminal object in $Cone(f,g)$, where $f:A\rightarrow C$, $g:B \rightarrow C$, i.e. there is a unique arrow from any cone to this cone.
Now if we apply the powerset functor, we get
$$
\begin{array}{ccc}
2^D & \rightarrow & 2^B \\
\downarrow & & \downarrow \\
2^A & \rightarrow & 2^C
\end{array}
$$
which seems to be commutative, but how to show that there is no uniqueness at this time?
And why doesn't it preserve pullbacks (only weakly)? Could you please give me a counterexample if such exists? 
 A: (I’m just giving partial hints, not full answers; let me know if these aren’t enough, and I can give more details.)
You ask “how to show there is no uniqueness?”  However, to show that $2^{(-)}$ weakly preserves pullbacks, you don’t need to show that there’s no uniqueness; the definition of a weak pullback doesn’t mention uniqueness one way or the other.
What you need to show is (by definition) just that the induced map $2^D \to 2^A \times_{2^C} 2^B$ is surjective.  In other words, given subsets $A' \subseteq A$ and $B' \subseteq B$ such that $f[A'] = g[B'] \subseteq C$, find some subset $D' \subseteq D$ whose images under projection are $A'$ and $B'$.
To give an example where non-uniqueness can arise, you just need two subsets $D_1', D_2'$ which are not the same, but whose images under projection both to $A$ and to $B$ are the same.  Often with pullbacks, it’s worth looking for examples in the special case of products; and in this case that suffices: there’s an example of non-uniqueness with $A = B = \{0,1\}$, $C = 1$, $D = \{0,1\}^2$.
