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• CommentRowNumber1.
• CommentAuthorxelxebar
• CommentTimeNov 26th 2010

Hello,

This is my first post here at nForum, and I’m a bit unsure whether my question is appropriate here, so please let me know if it should be moved or deleted. However, I’m just beginning to study Category Theory on my own, using Adamek, Herrlich and Strecker’s “Abstract and Concrete Categories”. I’m having a bit of a stumbling block with one of the exercises and would like to discuss it with someone.

The exercise just asks to show that $Set \ncong Set_*$ where $Set_*$ is the category of pointed sets and $\cong$ is equivalence of categories. Could somebody provide me with a little direction? I’d rather work from a hint than receive an answer, but I know that’s sometimes difficult for these elementary problems.

• CommentRowNumber2.
• CommentAuthorDavidRoberts
• CommentTimeNov 26th 2010

Think about the ’obvious’ functor $Set_* \to Set$, and see if it satisfies the criteria for a functor to be an equivalence of categories. I would link to equivalence of categories, but it takes a very sophisticated, n-categorical look at the concept, so I don’t recommend it.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeNov 26th 2010

but it takes a very sophisticated, n-categorical look

I’d say what makes this entry look sophisticated is not its $n$-categorical attitude, but that it pays so much attention to the axiom of choice.

I don’t recommend it.

Hopefully then somebody finds the time to improve it!

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeNov 26th 2010

Hopefully then somebody finds the time to improve it!

Okay, I did it myself. I reworked the beginning of equivalence of categories. And stated more prominently the central theorem that David wanted to point out to xelxebar, that if the axiom of choice does hold, then a functor is part of an equivalence precisely if it is essentially surjective and full and faithful.

• CommentRowNumber5.
• CommentAuthorTobyBartels
• CommentTimeNov 26th 2010
• (edited Nov 26th 2010)

if the axiom of choice does hold, then a functor is part of an equivalence precisely if it is essentially surjective and full and faithful

But that doesn’t depend on the axiom of choice at all! It’s just that people who believe in choice use a different definition, which is then wrong in the absence of choice.

The old version of the article was more or less even-handed, but if you’re going to rewrite it to single out a particular definition, then let’s make it the right one.

• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeNov 26th 2010

@xelxebar: think a bit about initial and terminal objects in each of those categories. (And I wouldn’t worry much at the beginning about all this talk of the axiom of choice, not at this point of your education. You can use the definition that involves two functors and two natural isomorphisms.)

• CommentRowNumber7.
• CommentAuthorTobyBartels
• CommentTimeNov 26th 2010

I wrote:

It’s just that people who believe in choice use a different definition, which is then wrong in the absence of choice.

Actually, on second thought, that’s not really true; Urs’s definition is fine even in the absence of choice, as long as you define ‘functor’ correctly. So I’ll put the article back more like how he had it.

@ xelxebar

Todd is right; don’t worry at all about this axiom-of-choice business. Your problem is much simpler than any of that. Take any characterisation of equivalence of categories and prove that an equivalence must map initial objects to initial objects and terminal objects to terminal objects. Then combine this with Todd’s suggestion, and you should have your answer.

• CommentRowNumber8.
• CommentAuthorDavid_Corfield
• CommentTimeNov 26th 2010
• (edited Nov 26th 2010)

Perhaps xelxebar might think about initial and terminal objects in the two categories. Then think about why the possession of zero objects must be preserved under equivalence.

Edit: I see similar advice came in while writing the above. Must be good advice then.

• CommentRowNumber9.
• CommentAuthorxelxebar
• CommentTimeDec 1st 2010

Thank you very much for your advice. It got me on the right track pretty quickly.