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Putting together §7.3 Theorem 1 and §8.3 Theorem 1 of Goldblatt Topoi we know that, in a topos where $[\top, \bot]: 1 \oplus 1 \to \Omega$ is an isomorphism, $\neg\circ\neg = 1_\Omega$.

Goldblatt's proof goes via facts about lattices and Boolean algebras. Shouldn't there be a simpler, more direct, proof that doesn't involve considering conjunctions/disjunctions, intersections/unions? But what is it?

(Sorry if this is well known and I have just searched badly. I had guessed that we could easily(?) show that in any topos (?) there is a pullback square as in question Showing that, in a topos where $\Omega \cong 1 \oplus 1$, then $\bot = \neg \top$ in a pullback square which could be pasted with the pullback defining $\neg$ to then show $\neg\circ\neg = 1_\Omega$. But I guessed wrong!)

Late starter
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    What is $\otimes$ here? – Captain Lama Feb 21 '25 at 16:22
  • Ooops. $\oplus$ for the object in a coproduct. – Late starter Feb 21 '25 at 16:42
  • Note that if you have this, then the proof of https://math.stackexchange.com/questions/5037257/showing-that-in-a-topos-where-omega-cong-1-oplus-1-then-bot-neg-top is trivial (because it implies that $\neg$ is an isomorphism and it is trivial to describe the pullback along an isomorphism). – Zufallskonstante Feb 21 '25 at 16:43
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    @Zufallskonstante Certainly if we already have a proof that when $\Omega \cong 1 \oplus 1$ then $\neg\neg = 1_\Omega$. – Late starter Feb 21 '25 at 17:04
  • Is the assumption that there exists some isomorphism $\Omega \simeq 1 \sqcup 1$, or that the specific morphism $\begin{bmatrix} \top & \bot \end{bmatrix} : 1 \sqcup 1 \to \Omega$ is an isomorphism? – Daniel Schepler Feb 21 '25 at 21:48
  • @DanielSchepler those are equivalent, see bullet 4 here – Naïm Camille Favier Feb 21 '25 at 22:07
  • @DanielSchepler Thanks -- yes at least in reference to Goldblatt it should be the second. I have edited to correct. – Late starter Feb 21 '25 at 22:09
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    (Though I suppose that there is a supplementary question -- can we get the conclusion that $\neg\neg = 1$ from just the unspecific assumption that there is an isomorphism, without going though the lemma that $[\top, \bot]$ in particular is iso?) – Late starter Feb 21 '25 at 22:19
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    In that case doesn't that imply $\lnot$ is just the transport under that isomorphism of the "component interchange" morphism $\begin{bmatrix} i_1 & i_2 \end{bmatrix} : 1 \sqcup 1 \to 1 \sqcup 1$, which it's trivial to show is self-inverse? – Daniel Schepler Feb 22 '25 at 14:54

1 Answers1

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In any topos, we have $\lnot \circ \top = \bot$ and $\lnot \circ \bot = \top$. Therefore, $\lnot \circ \lnot \circ \top = \top$ and $\lnot \circ \lnot \circ \bot = \bot$, from which we see that $\lnot \circ \lnot \circ \begin{bmatrix} \top & \bot \end{bmatrix} = \begin{bmatrix} \top & \bot \end{bmatrix}$. With the additional assumption that $\begin{bmatrix} \top & \bot \end{bmatrix}$ is an isomorphism, that implies that $\lnot \circ \lnot = 1_{\Omega}$.

Daniel Schepler
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