Timeline for Example of an unnatural isomorphism
Current License: CC BY-SA 3.0
15 events
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| Feb 9, 2015 at 4:17 | comment | added | Exterior | @MartinBrandenburg what is meant by "conjugation"? | |
| Aug 18, 2013 at 6:51 | comment | added | Martin Brandenburg | I have now added the most basic example which one can think of and placed it as example 1. In the comments above, example 1 refers to what is now example 2. | |
| Aug 18, 2013 at 6:50 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Aug 16, 2013 at 10:42 | comment | added | Omar Antolín-Camarena | That's right, @TomLeinster! Thanks for catching the mistake. | |
| Aug 16, 2013 at 6:40 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Aug 16, 2013 at 6:35 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Aug 16, 2013 at 0:29 | comment | added | Tom Leinster | Omar, I don't think you said what you meant to say. If what you said was true, we could compose your natural iso with the first-projection functor $C \times C \to C$ to obtain a natural iso between Ord and Sym - and that's impossible. I guess you actually meant to say that there's a natural iso between the two endofunctors of $C$ defined by $X \mapsto \text{Ord}(X) \times \text{Sym}(X)$ and $X \mapsto \text{Ord}(X) \times \text{Ord}(X)$. | |
| Aug 15, 2013 at 14:37 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Aug 15, 2013 at 2:11 | history | made wiki | Post Made Community Wiki by Ben Webster♦ | ||
| Aug 14, 2013 at 18:39 | comment | added | Eric Wofsey | The first example is perhaps more transparent if you consider it as a special case of the second example, where $G=S_n$ and you let it act on itself either by translation (to get Ord) or conjugation (to get Sym). | |
| Aug 14, 2013 at 17:39 | comment | added | ACL | @OmarAntolín-Camarena: Indeed, if $G$ is a group and $X$ is a principal homogeneous space under $G$, then $G\times X$ is naturally isomorphic to $X\times X$. | |
| Aug 14, 2013 at 17:33 | comment | added | Omar Antolín-Camarena | As a bonus exercise, it is fun to show that $(\mathrm{Ord},\mathrm{Sym})$ is naturally isomorphic to $(\mathrm{Ord},\mathrm{Ord})$ (Here, $(F,G)$ denotes the functor $C \to C \times C$ given by $(F,G)(X)=(FX,GX)$, $(F,G)(f)=(Ff,Gf)$). | |
| Aug 14, 2013 at 15:15 | comment | added | Omar Antolín-Camarena | By the way, example 1 is well-known in combinatorics, there people say "the species of permutations and the species of total orders have (1) the same exponential generating functions, but (2) different isomorphism type generating functions". (1) says that $\mathrm{Sym}(X) \cong \mathrm{Ord}(X)$ for all $X$, (2) shows that the functors can't be isomorphic. | |
| Aug 14, 2013 at 12:00 | comment | added | Karol Szumiło | I think that this answer is severely underappreciated. Personally, I consider the first example here to be the most illuminating among all the examples mentioned in the answers to this question. It's a very useful exercise to explicitly write down both functors for some small sets and analyze them. The non-existence of a natural isomorphism between these functors really helps me appreciate what naturality is all about. | |
| Aug 14, 2013 at 8:51 | history | answered | Martin Brandenburg | CC BY-SA 3.0 |