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1. • CommentRowNumber1.
   • CommentAuthorYimingXu
   • CommentTimeMar 2nd 2022
   • (edited Mar 2nd 2022)
   • PermaLink
   Author: YimingXu
   Format: MarkdownItexHi, all. I am confused by the following sentence from the document [here](https://www.jstor.org/stable/2275472?casa_token=hi4gsrP5DHMAAAAA:l5929tBx2JWhyJS9B1aoR8iB3AaD1o8NByGOmoj2MHTMoI-LrnB4EzfRSa-si01RyxIz4dHbuldXesD-x9LvW-TmmojxoTMS7XbUrNWEsykCsx0p) In page 1249, The paper says: > By Theorem 3, $2^{op}\cong 2$ Theorem 3 says "any generator with exactly three endomorphisms is isomorphic to $2$", therefore, to show $2^{op}\cong 2$, there are two things to check: 1. $2^{op}$ is a generator, 2. there are exactly three functors $2^{op} \to 2^{op}$. I agree that the axioms $(A^{op})^{op} = A$ and $(F^{op})^{op} = F$ makes sure that there is a one-to-one correspondence from functors $A \to B$ to functors $A^{op} \to B^{op}$. My confusion is to show the first point, that is, why is $2^{op}$ a generator? That is, given functors $f,g:A\to B$ such that $ f \neq g$, how can we come up with a functor $a:2^{op} \to A$ such that $f\circ a \neq g \circ a$? The author does not say anything like "we have an isomorphism $2\to 2^{op}$", and the paper does not assume any "extensionality of categories" (i.e., we are not able to say two categories are equal iff they have equal collection of objects and a equal collection of arrows), therefore, the assumption $(A^{op})^{op} = A$ does not really say anything about the objects an arrows of $A^{op}$, since the operation is applied on $A$, not on its object (which are **defined** to be functors $1\to A$), and not on its arrows (which are **defined** to be functors $2\to A$). Also note that this sentence is stated before assuming $1 = 1^{op}$ and $2 = 2^{op}$, instead, the workflow seems to be: the axioms are sufficient to prove $2\cong 2^{op}$, and since assuming $2 = 2^{op}$ does not break consistency, we can add this assumption. Or is the author sloppy here with the "minor detail"? If so, I think I am in trouble. Any hint to fix this point? Thanks to any attempt to help!

   Hi, all.

   I am confused by the following sentence from the document here

   In page 1249, The paper says:

   By Theorem 3, $2^{op}\cong 2$

   Theorem 3 says “any generator with exactly three endomorphisms is isomorphic to $2$”, therefore, to show $2^{op}\cong 2$, there are two things to check: 1. $2^{op}$ is a generator, 2. there are exactly three functors $2^{op} \to 2^{op}$.

   I agree that the axioms $(A^{op})^{op} = A$ and $(F^{op})^{op} = F$ makes sure that there is a one-to-one correspondence from functors $A \to B$ to functors $A^{op} \to B^{op}$. My confusion is to show the first point, that is, why is $2^{op}$ a generator? That is, given functors $f,g:A\to B$ such that $f \neq g$, how can we come up with a functor $a:2^{op} \to A$ such that $f\circ a \neq g \circ a$?

   The author does not say anything like “we have an isomorphism $2\to 2^{op}$”, and the paper does not assume any “extensionality of categories” (i.e., we are not able to say two categories are equal iff they have equal collection of objects and a equal collection of arrows), therefore, the assumption $(A^{op})^{op} = A$ does not really say anything about the objects an arrows of $A^{op}$, since the operation is applied on $A$, not on its object (which are defined to be functors $1\to A$), and not on its arrows (which are defined to be functors $2\to A$). Also note that this sentence is stated before assuming $1 = 1^{op}$ and $2 = 2^{op}$, instead, the workflow seems to be: the axioms are sufficient to prove $2\cong 2^{op}$, and since assuming $2 = 2^{op}$ does not break consistency, we can add this assumption.

   Or is the author sloppy here with the “minor detail”? If so, I think I am in trouble. Any hint to fix this point?

   Thanks to any attempt to help!

2. • CommentRowNumber2.
   • CommentAuthorHurkyl
   • CommentTimeMar 2nd 2022
   • (edited Mar 2nd 2022)
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   Author: Hurkyl
   Format: MarkdownItex(deleted)

   (deleted)

3. • CommentRowNumber3.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 4th 2022
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   Author: DavidRoberts
   Format: MarkdownItexWhy doesn't taking the opposite of the arrow $2\to A^{op}$ that shows $f^{op}\neq g^{op}$ work? Since $f\neq g$, we have $f^{op}\neq g^{op}$, since $(-)^{op}$ is an isomorphism on hom-sets, after all.

   Why doesn’t taking the opposite of the arrow $2\to A^{op}$ that shows $f^{op}\neq g^{op}$ work? Since $f\neq g$, we have $f^{op}\neq g^{op}$, since $(-)^{op}$ is an isomorphism on hom-sets, after all.

4. • CommentRowNumber4.
   • CommentAuthorYimingXu
   • CommentTimeMar 4th 2022
   • (edited Mar 4th 2022)
   • PermaLink
   Author: YimingXu
   Format: MarkdownItex>Why doesn't taking the opposite of the arrow $2\to A^{op}$ that shows $f^{op}\neq g^{op}$ work? Since $f\neq g$, we have $f^{op}\neq g^{op}$, since $(-)^{op}$ is an isomorphism on hom-sets, after all. Thank you very much! I thought too much about comprehension, that was my stupid. I have checked that does work so I am out of trouble!

   Why doesn’t taking the opposite of the arrow $2\to A^{op}$ that shows $f^{op}\neq g^{op}$ work? Since $f\neq g$, we have $f^{op}\neq g^{op}$, since $(-)^{op}$ is an isomorphism on hom-sets, after all.

   Thank you very much! I thought too much about comprehension, that was my stupid. I have checked that does work so I am out of trouble!