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Following on from this question of mine from a long time ago, I would like to know

What is the lattice of truth values in $\mathbf{Set}^2$?

Which lattice?

I mean the Heyting algebra.

I suspect that $(1,1)$ is at the top and $(0,0)$ is at the bottom. Since I think $(0,1)$ and $(1,0)$ are isomorphic in $\mathbf{Set}^2$, I suspect that they occupy the same space in the lattice; and that would be in the middle, right? So the lattice is a straight line . . .

What is making me doubt this?

The answer to the question above states that, at least in Goldbatt's book, we don't need to consider such isomorphisms. Was that misleading of them to say?

Shaun
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  • But aren't $(0,0)$ and $(1,1)$ isomorphic, too? – Shaun Nov 03 '24 at 19:28
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    What would the isomorphism in $\mathbf{Set}^2$ between $(0, 1)$, and $(1, 0)$ be? And for your comment, what would the isomorphism in $\mathbf{Set}^2$ between $(0, 0)$ and $(1, 1)$ be? – Daniel Schepler Nov 03 '24 at 19:37
  • I can't answer that, @DanielSchepler, without making an error somewhere, apparently; I don't know where though. They're coordinates; swap them in the first case and relabel them in the second. – Shaun Nov 03 '24 at 19:43
  • I don't get the downvote. Is this a stupid question somehow? – Shaun Nov 03 '24 at 19:47

1 Answers1

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Recall that the morphisms in $\mathbf{Set}^2$ are defined such that $$\operatorname{Hom}_{\mathbf{Set}^2}((A, B), (C, D)) \overset{def}{=} \operatorname{Hom}_{\mathbf{Set}}(A, C) \times \operatorname{Hom}_{\mathbf{Set}}(B, D).$$

Therefore, there is no morphism at all between $(0, 1)$ and $(1, 0)$ in either direction, much less an isomorphism in $\mathbf{Set}^2$, much less an isomorphism as subobjects of $(1, 1)$.

Similarly, there is no morphism at all from $(1, 1)$ to $(0, 0)$, much less an isomorphism in $\mathbf{Set}^2$, much less an isomorphism as subobjects of $(1, 1)$.

Therefore, the global sections of $\Omega$ are isomorphic, as a lattice, to $\{ 0, 1 \}^2$. In particular, this is a boolean algebra (and as a matter of fact, $\mathbf{Set}^2$ is a boolean topos).

In fact, $\Omega$ as an object of $\mathbf{Set}^2$ is isomorphic to $(2, 2)$ -- which agrees with the previous statement since the global sections of $(2, 2)$ are $$\operatorname{Hom}_{\mathbf{Set}^2}((1, 1), (2, 2)) \simeq 2 \times 2.$$

Daniel Schepler
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