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1. • CommentRowNumber1.
   • CommentAuthorYimingXu
   • CommentTimeNov 15th 2021
   • (edited Nov 15th 2021)
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   Author: YimingXu
   Format: MarkdownItexHi, all. I am reading the SEAR nlab page here: https://ncatlab.org/nlab/show/SEAR Regarding Theorem 2.8, it says: Theorem 2.8. For any sets A and B, $A\times B$ is a product of A and B in the category Set, and a coproduct in the category Rel. and the proofs says just in naive set theory. But the coproduct of Rel in naive set theory is the disjoint union! I am confused how can $A\times B$ can serve as the coproduct of Rel, I know that the coproduct/product in Rel is the disjoint union, but how can the cartesian product coincide with the disjoint union? Given relation $f:A\to X,g:B\to X$, how can we define the relation $A \times B \to X$ so it is a coproduct? I assume the inclusion relations $A\to A \times B$ and $B\to A\times B$ is $(a,(a1,b1))$ holds iff $a = a1$, and $(b,(a1,b1))$ holds iff $b = b1$. I may also make mistake here but I still wonder what else could it be. Thank you for any explaination! (and sorry if it turns out to be a stupid question...)

   Hi, all.

   I am reading the SEAR nlab page here: https://ncatlab.org/nlab/show/SEAR

   Regarding Theorem 2.8, it says:

   Theorem 2.8. For any sets A and B, $A\times B$ is a product of A and B in the category Set, and a coproduct in the category Rel.

   and the proofs says just in naive set theory.

   But the coproduct of Rel in naive set theory is the disjoint union! I am confused how can $A\times B$ can serve as the coproduct of Rel, I know that the coproduct/product in Rel is the disjoint union, but how can the cartesian product coincide with the disjoint union? Given relation $f:A\to X,g:B\to X$, how can we define the relation $A \times B \to X$ so it is a coproduct?

   I assume the inclusion relations $A\to A \times B$ and $B\to A\times B$ is $(a,(a1,b1))$ holds iff $a = a1$, and $(b,(a1,b1))$ holds iff $b = b1$. I may also make mistake here but I still wonder what else could it be.

   Thank you for any explaination! (and sorry if it turns out to be a stupid question…)

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeNov 15th 2021
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   Author: Urs
   Format: MarkdownItexNot to answer your question (i have never worked on or even read that page) but just to note that in order to hyperlink to an entry here, just enclose its name in double square brackets: `[[SEAR]]` produces *[[SEAR]]*. Specifically, you are asking about [this Prop.](https://ncatlab.org/nlab/show/SEAR#product). Hm, but looking at what it says there, I agree that it *sounds* like directly contradicting both what one expects as well as what it says at *[[Rel]]* ([here](https://ncatlab.org/nlab/show/Rel#LimitsAndColimit)). Maybe a typo.

   Not to answer your question (i have never worked on or even read that page) but just to note that in order to hyperlink to an entry here, just enclose its name in double square brackets: [[SEAR]] produces SEAR. Specifically, you are asking about this Prop..

   Hm, but looking at what it says there, I agree that it sounds like directly contradicting both what one expects as well as what it says at Rel (here). Maybe a typo.

3. • CommentRowNumber3.
   • CommentAuthorYimingXu
   • CommentTimeNov 15th 2021
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   Author: YimingXu
   Format: MarkdownItexThanks for the quick reply and comment anyway, and thank you for teaching me to write the link. By the way, I think I can report a typo now in a relative page [[allegory]], in the subsection [[allegory#division-allegory]], I saw: That is: given $r:A\to B$ and $s:A\to C$, there exists $r/s:B\to C$ such that $t\le s/r\in hom(B,C)$... It seems that the first $r/s$ should be $s/r$.

   Thanks for the quick reply and comment anyway, and thank you for teaching me to write the link.

   By the way, I think I can report a typo now in a relative page allegory, in the subsection allegory#division-allegory, I saw:

   That is: given $r:A\to B$ and $s:A\to C$, there exists $r/s:B\to C$ such that $t\le s/r\in hom(B,C)$…

   It seems that the first $r/s$ should be $s/r$.

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeNov 15th 2021
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   Author: Urs
   Format: MarkdownItexIf you spot typos and think you know what you are doing, then you are doing the community a favor by going ahead and fixing them. Thanks!

   If you spot typos and think you know what you are doing, then you are doing the community a favor by going ahead and fixing them. Thanks!

5. • CommentRowNumber5.
   • CommentAuthorGuest
   • CommentTimeNov 15th 2021
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   Author: Guest
   Format: MarkdownItexRe #1: For a non-naïve set theory proof that [theorem 2.8](https://ncatlab.org/nlab/revision/SEAR/55#product) is wrong: [[Rel]] is the [[Kleisli category]] of the power set monad, so the inclusion $\mathrm{Set}\to\mathrm{Rel}$ [[monad#RelationToAdjunctionsAndMonadicity|preserves colimits]], but the product set $A\times B$ and disjoint union set $A\sqcup B$ are generally not isomorphic in Rel since they needn’t have the same cardinality (and invertible relations are [[bicategory of maps|functions]]). (That $A\sqcup B$ is also the product in Rel follows from $\operatorname{Rel}(X,A\sqcup B) \cong\operatorname{Set}(X\times (A\sqcup B),\Omega) \cong\operatorname{Set}((X\times A)\sqcup (X\times B),\Omega) \cong\operatorname{Set}(X\times A,\Omega)\times\operatorname{Set}(X\times B,\Omega) \cong\operatorname{Rel}(X,A)\times\operatorname{Rel}(X,B)$.) The disjoint union of sets isn’t introduced until later on in the article, so I’ll just remove the mention of Rel from the theorem.

   Re #1: For a non-naïve set theory proof that theorem 2.8 is wrong: Rel is the Kleisli category of the power set monad, so the inclusion $\mathrm{Set}\to\mathrm{Rel}$ preserves colimits, but the product set $A\times B$ and disjoint union set $A\sqcup B$ are generally not isomorphic in Rel since they needn’t have the same cardinality (and invertible relations are functions). (That $A\sqcup B$ is also the product in Rel follows from $\operatorname{Rel}(X,A\sqcup B) \cong\operatorname{Set}(X\times (A\sqcup B),\Omega) \cong\operatorname{Set}((X\times A)\sqcup (X\times B),\Omega) \cong\operatorname{Set}(X\times A,\Omega)\times\operatorname{Set}(X\times B,\Omega) \cong\operatorname{Rel}(X,A)\times\operatorname{Rel}(X,B)$.)

   The disjoint union of sets isn’t introduced until later on in the article, so I’ll just remove the mention of Rel from the theorem.

6. • CommentRowNumber6.
   • CommentAuthorYimingXu
   • CommentTimeNov 19th 2021
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   Author: YimingXu
   Format: MarkdownItexThank you, then I will just skip it. And thanks for the explanation above! But I am still interested in what is the intended theorem there. If someone knows, thanks for telling me or just editing the page...

   Thank you, then I will just skip it. And thanks for the explanation above!

   But I am still interested in what is the intended theorem there. If someone knows, thanks for telling me or just editing the page…

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeNov 19th 2021
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   Author: Mike Shulman
   Format: MarkdownItexI expect this was just a mistake. Whoever wrote that (probably me) may have been thinking of the fact that, as mentioned above, the coproduct in Set is both a product and a coproduct in Rel, and mistakenly dualized it.

   I expect this was just a mistake. Whoever wrote that (probably me) may have been thinking of the fact that, as mentioned above, the coproduct in Set is both a product and a coproduct in Rel, and mistakenly dualized it.