Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

• Forgot your password?
• Apply for membership

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,

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

   I agree that the axioms and makes sure that there is a one-to-one correspondence from functors to functors . My confusion is to show the first point, that is, why is a generator? That is, given functors such that , how can we come up with a functor such that ?

   The author does not say anything like “we have an isomorphism ”, 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 does not really say anything about the objects an arrows of , since the operation is applied on , not on its object (which are defined to be functors ), and not on its arrows (which are defined to be functors ). Also note that this sentence is stated before assuming and , instead, the workflow seems to be: the axioms are sufficient to prove , and since assuming 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)
   • PermaLink
   Author: Hurkyl
   Format: MarkdownItex(deleted)

   (deleted)

3. • CommentRowNumber3.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 4th 2022
   • PermaLink
   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 that shows work? Since , we have , since 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 that shows work? Since , we have , since 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!