Why doesn't the functor $\bar{\mathcal{P}}\bar{\mathcal{P}}$ preserve pullbacks?

Question

I've tried finding examples on my own but the sizes of the sets is a bit hard to manage. In the litterature I've seen this fact referenced in a few places but they all point to Rutten: Universal coalgebra: a theory of systems which mentions, in the very paragraph, an example where $\bar{\mathcal{P}}\bar{\mathcal{P}}$ supposedly does not preserve pullbacks. Recall that $\bar{\mathcal{P}}\bar{\mathcal{P}}$ is defined as:

$$\bar{\mathcal{P}}\bar{\mathcal{P}}(f):\bar{\mathcal{P}}\bar{\mathcal{P}}(A)\to\bar{\mathcal{P}}\bar{\mathcal{P}}(B) \\ Y\mapsto \{X\subseteq B\mid f^{-1}[X]\in Y\}$$

for all $Y\subseteq A$ and $f[X]=\{f(x)\mid x\in X\}$.

"There is one functor in our list above that does not even preserve weak pullbacks. It is the contravariant powerset functor composed with itself $ \bar{\mathcal{P}}\circ\bar{\mathcal{P}}$ .

Take, for instance, $S = \{s_1, s_2 , s_3 \}$; $T = \{t_1 , t_2 , t_3 \}$; U = $\{u_1 , u_2 \}$; $f : S → U$ denoted by $\{s_1 \to u_1 , s_2 \to u_1 , s_3 \to u_2 \}$ and $g : T → U$ denoted by $\{t_1 \to u_1 ,t_2 \to u_2, t_3 \to u_2 \}$. Then the image of the pullback of f and g is not a pullback and not even a weak pullback."

The pullback I think works is $W=\{(s_1,t_1),(s_2,t_1),(s_3,t_3),(s_3,t_2) \}$ but it's intractable for me to find the image of the coalgebra map of this set. There must be some other way than calculating 3*256+16 elements and checking each one.

Answer 1

I think that this is just a cardinality argument. $W$ is the pullback in sets, and has four elements. Applying the powerset functor ($A\mapsto 2^A$) twice yields a $2^{2^4} = 2^{16}$ element set. Since the pullback of $2^{2^f}$ with $2^{2^g}$ is a subset of $2^{2^S}\times 2^{2^T}$, which has $2^{2^3}\cdot 2^{2^3} = 2^8\cdot 2^8 = 2^{16}$ elements, we can say immediately that pullbacks won't be preserved as soon as there is a pair $(x,y)\in 2^{2^S}\times 2^{2^T}$ with $2^{2^f}(x) \neq 2^{2^g}(y)$. Since (for example) the function $2^{2^f}$ is not constant, this is indeed the case.