Preservation of weak pullback

Question

A weak pullback is defined in the same way as a pullback, but the arrow to the vertex of the limit cone is not required to be unique.

Here's the problem:

Let $\mathscr P:\mathbf {Set}\to\mathbf{Set}$ be the powerset functor.

(a) Does $\mathscr P$ preserve pullbacks?

(b) Does $\mathscr P$ preserve weak pullbacks?

(a) I suppose the answer is not. Consider the diagram on the left:

The triple $(L,{\rm proj}_1,{\rm proj}_2)$ is a pullback. Its image is shown on the right. On the right picture, the elements (which happen to be sets) that are circled in green, get mapped to $\{0\}$ in the lower right corner and the "empty sets" get mapped to the empty set in the lower right corner. So by the general description of pullbacks in $\mathbf{Set}$, we conclude that the vertex of the limit cone is $\{(\{1\},\{1\}),(\{1\},\{2\}),(\{1\},\{2,3\}),(\{2\},\{1\}),(\{2\},\{2\}),(\{2\},\{2,3\}),(\{2,3\},\{1\}),(\{2,3\},\{2\}),(\{2,3\},\{2,3\})\}$.

It has $9$ elements whereas $\mathscr P(L)$ has 16 elements. Since any two limits are isomorphic, $\mathscr P(L)$ cannot be a limit. So $\mathscr P$ does not preserve limits. Is that right?

(b) According to Powerset functor weakly preserves pullbacks. this statement is true, but I don't understand the answer given there (though I stole the idea at the end and used it in (a) above). It seems the author of the answer is using another definition of weak pullback, and I don't understand why what he is claiming is enough. How to use my definition directly to prove that weak pullbacks are preserved?

Answer 1

Essentially you want proof of the following

Theorem In the category of sets if a map $$g:N \twoheadrightarrow L$$ is surjective and $L$ is a pullback (for some diagram) then we have that $N$ is a weak-pullback (for the same diagram).

and I am assuming that your definition of a weak-pullback is that it satisfies the "$(\forall X \text{ making the diagram commute)} (\exists$ an arrow to $N$)" as opposed "$(\forall X \text{ making the diagram commute)} \exists!$ arrow to $N$," or as you said "but the arrow to the vertex of the limit cone is not required to be unique" where the diagram is the following
$\require{AMScd}$ \begin{CD} L @>p_1>>B\\ @VV{p_2}V @VV{f_2}V\\ A @>{f_1}>> C \end{CD}

I give two explanations

1. Informal explanation
2. Formal proof

Informal explanation

If we assume the axiom of choice then sets are isomorphic if they have the same number of objects i.e. same cardinality. Universal properties/objects/maps are "optimal" solutions to problems in the sense that they are the exact minimum of properties needed to satisfy a definition (i.e. they capture the essence of the definition); we could conjecture that since sets are nothing more than a "cardinality" then a set with more elements than the pullback will "satisfy the same properties (and more)," a set with fewer elements than the pullback will "fail to satisfy some of the properties," and finally a set with the same number of objects as the pullback will satisfy exactly the same properties and nothing more. Here we mean properties related to the pullback such as $p_1(x)=a,$ $p_2(x)=b,$ and $f_1(p_1(x))=f_2(p_2(x)$ for each $a,b$ such that $f_1(a) = f_2(b)$.

Let us try to use this intuition to provide a formal proof.

Proof (Assuming Axiom of Choice)

Suppose that there exists a function $g:N \twoheadrightarrow L$ that is surjective. Define $h_1:N \longrightarrow A$ and $h_2:N \longrightarrow B$ as follows: let $h_1(n) = p_1(g(n))$ and $h_2(n) = p_2(g(n))$. One can easily check that $(N,h_1,h_2)$ satifies the property "$(\forall (X,q_1,q_1) \text{ making the diagram commute)} (\exists$ an arrow to $N$);" this is because $g$ is surjective. Indeed, because "$p_1,p_2$ are the pullback" we have that $\exists! \psi:X \longrightarrow L$ that makes the diagram commute; in order to define a $\varphi:X \longrightarrow N$ just use the axiom of choice to set $\varphi(x)$ equal to any $n \in g^{-1}(\psi(x))$. This is well-defined because $g$ is surjective. QED

Remark The reason why it is versal and not universal (for an explanation of the name versal see this book by Yuri Manin) is that the extra elements in $N$ allow for a lot of freedom in how to define $\varphi$.

We are finally to prove the (main) question you asked.

Main Theorem The power set functor preserves weak pullbacks.

Notice that if $L$ is the pullback in the following diagram

$\require{AMScd}$ \begin{CD} L @>p_1>>B\\ @VV{p_2}V @VV{f_2}V\\ A @>{f_1}>> C \end{CD}

then it is straightforward to prove that

$\require{AMScd}$ \begin{CD} 2^L @>\mathcal{P}(p_1)>>2^B\\ @VV{\mathcal{P}(p_2)}V @VV{\mathcal{P}(f_2)}V\\ 2^A @>{\mathcal{P}(f_1)}>> 2^C \end{CD}

commutes (here the $\mathcal{P}(p_1),\mathcal{P}(p_2)$ play the role of the $h_1,h_2$ in the previous proof). By the previous theorem, it suffices to prove that

$$\exists g: 2^L \twoheadrightarrow 2^A\times_{2^C}2^B$$.

Because $L \cong \{(a,b) \in A\times B \ | \ f_1(a) = f_2(b)\}$ and $2^A\times_{2^C}2^B \cong \{(A',B') \in \mathcal{P}(A)\times \mathcal{P}(B) \ | \ f_1(A') = f_2(B')\}$ we can define $g: 2^L \twoheadrightarrow 2^A\times_{2^C}2^B$ as follows

$$g: \{(a_\lambda,b_{\lambda}) \ | \lambda \in \Lambda \ \} \mapsto (\{a_\lambda \ | \lambda \in \Lambda \ \}, \{b_{\lambda} \ | \lambda \in \Lambda \ \}) $$

Suppose that $f_1(A') = f_2(B')$ then we have that for each $a \in A'$ there exists $b \in B'$ such that $f_1(a) = f_2(b)$ so that there exists by the definition of $L$ some $ B'' \subset B'$ such that $A'\times B'' \subset L$ and likewise that for each $b \in B'$ there exists $a \in A'$ such that $f_2(b) = f_1(a)$ so that there exists by the definition of $L$ some $ A'' \subset A'$ such that $A''\times B' \subset L$. Therefore $(A'\times B'')\cup(A''\times B') \subset L$ and $g((A'\times B'')\cup(A''\times B')) = (A', B')$ by definition; therefore $g$ is surjective. QED