Preservation of weak pullback 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?
 A: 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 


*

*Informal explanation

*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
