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?

  • $\begingroup$ WHat do you mean by «it seems to be commutative»? Either the square commutes or it doesn't, and it is not really difficult to check which option is the correct one. Have you tried? $\endgroup$ – Mariano Suárez-Álvarez Feb 3 '14 at 5:50

(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 $A = B = \{0,1\}$, $C = 1$, $D = \{0,1\}^2$.

  • $\begingroup$ But why such a map must be surjective? By the definition we have, say, in cones-notation, that the weak pullback is such an object in $Cone(f,g)$ that there is an arrow (not necessarily unique) from any cone to this one. Sorry, could you please give some more details? As far as I understood your answer, we want to show that if the first diagram is a pullback diagram, then the second one is a weak pullback diagram, i.e. there is a map to the actual pullback (which exists, as we live in $Set$). Is it right? $\endgroup$ – AlexCon Dec 8 '13 at 21:08
  • $\begingroup$ I’m using the fact that in $\mathbf{Set}$, a cone over $X,Y,Z$ with vertex $W$ is a weak pullback if and only if the induced map $W \to X \times_Z Y$ is surjective. (This is sometimes used as the definition of a weak pullback, in $\mathbf{Set}$.) As you say, what you need to show is that assuming the original square $D,A,B,C$ is a pullback, then the square $2^D,2^A,2^B,2^C$ is a weak pullback. But this means not just showing that the map to the actual pullback exists (which, as you say, is automatically true, by the universal property of the actual pullback) but that it’s surjective. $\endgroup$ – Peter LeFanu Lumsdaine Dec 9 '13 at 14:54
  • $\begingroup$ Can you please suggest some resources like books, where the information about weak pullbacks can be found (especially, this surjective criterion)? $\endgroup$ – AlexCon Dec 10 '13 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.