4

$\require{AMScd}$Suppose a topos is classical/Boolean in the sense that $\Omega$ is isomorphic to $1 \oplus 1$.

Then (if I'm following) this square should commute and in fact be a pullback

$$ \begin{CD} 1 @>!>> 1 \\ @V{\top}VV @VV{\bot}V \\ \Omega @>>{\neg}> \Omega\end{CD}$$

I am reading Goldbatt, Topoi, and §8.3 Theorem 1 says that the square commutes. But here and in §7.3 everything seems related to facts about lattices and Boolean algebras. Shouldn't there be a simpler, more direct, proof that the diagram is a pullback in a classical topos (i.e. a proof that doesn't involve considering conjunctions/disjunctions, intersections/unions)? I am struggling to find one. Help please!

Late starter
  • 242

1 Answers1

3

To show that the square commutes we need to prove $\neg \circ \top = \bot$.

Consider this diagram: $$ \begin{CD} 0@>0_1>>1 @>!_1>> 1 \\ @V{!_0}VV@V{\bot}VV @VV{\top}V \\ 1 @>>\top>\Omega @>>{\neg}> \Omega\end{CD}$$

The right square is the pullback that defines $\neg$, as the characteristic arrow of $\bot$. The left square is a pullback that defines $\bot$, as the characteristic arrow of $0_1$, the unique arrow from the initial object. (The square is flipped diagonally.) By the pasting lemma, the whole rectangle is a pullback. The composition along the bottom must be the characteristic arrow of $!_0 = 0_1$, which is $\bot$, and we end up with the desired $\neg\circ \top = \bot$.

So the square commutes in any topos where $\bot$ exists (where arrows from the initial object are monic). We did not use $\Omega \simeq 1 + 1$ yet.

To show that this is in fact a pullback is a bit harder. We need the fact that toposes are extensive categories, and this is not entirely trivial to prove.

Specifically, in an extensive category, we have that for "double square" diagrams such as the one below, if the bottom row is a coprodcut, the top row is a coproduct iff both squares are pullbacks $$ \begin{CD} 1@>\top >> \Omega @<\bot<< 1\\ @V{1}VV@V{\neg}VV @VV{1}V \\ 1 @>>\bot>\Omega @<<{\top}< 1 \end{CD}$$ It is at this point where we use the assumption that $\Omega \simeq 1 + 1$, and, as a result, the left square is a pullback (which is the pullback you wanted). The right square is also a pullback, but that is just the definition of $\neg$.

It may be provable without the extensive property, if anyone knows, please comment.

Tony Dolezal
  • 354
  • How did I miss it.... We have both $\neg \circ \bot = \top$ (defn. of $\bot$) and the just proven $\neg \circ \top = \bot$, so plugging one into the other gives $\neg\circ \neg \circ\top = \top$ as well as $\neg\circ \neg \circ\bot = \bot$. In general, there could be more elements in $\Omega$ but in this case, $\neg\circ \neg$ and $1_\Omega$ agree on all (well, both) elements of $\Omega$, and since topos bivalence also gives extensionality, it must be indeed $\neg \circ \neg = 1_\Omega$. The extensive property is only needed to prove that complements exist. – Tony Dolezal Feb 20 '25 at 01:41
  • @TonyDolezal How do you get $\neg \circ \bot = \top$ by defn? Isn’t $\bot = \neg \circ \top$ the definition? Also, is it true that in General 1+1 has only two global elements? I am thinking here about the sheaf topos on a discrete space. – Zufallskonstante Feb 20 '25 at 06:02
  • @NaïmFavier I also had this argument in mind, but then I wasn’t so sure anymore if the classification of endomorphisms oft 1+1 is really so simple, again thinking about a sheaf topos on a discrete space. – Zufallskonstante Feb 20 '25 at 06:05
  • @Zufallskonstante you're right, that doesn't work. – Naïm Camille Favier Feb 20 '25 at 11:20
  • Right, my comment is wrong. I guess the extensivity used in my original proof is unavoidable. – Tony Dolezal Feb 20 '25 at 13:16
  • @Zufallskonstante I was also inclined to try the same argument that you had in mind, and then realised I couldn't nail down my assumption about the endomorphisms of 1 + 1. So I've tried asking a question here: https://math.stackexchange.com/questions/5037629/how-many-arrows-in-a-topos-from-1-coprod-1-to-1-coprod-1 – Peter Smith Feb 20 '25 at 16:19