Timeline for Nice examples of groups which are not obviously groups
Current License: CC BY-SA 3.0
7 events
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| Apr 18, 2013 at 1:31 | comment | added | user17945 | Let me try again: Ok, just looked up the definition. So to put my question in group object terms, one must have a sensible notion of $T 1:\ast\rightarrow TG$, where $\ast$ is a terminal object in the category of sets, and $1$ the unit map. In other words, you need to show an identity exists on $TG$. Again, this might be due to my naivete regarding category theory, but it just seems to me like you are begging the question somewhat, in that you're assuming this is already true. The existence of an identity for $TG$ wrt multiplication $Tm$ is essentially the proof of the result I mentioned. | |
| Apr 18, 2013 at 1:25 | comment | added | user17945 | Again, this might be due to my naivete regarding category theory, but it just seems to me like you are begging the question somewhat, in that you're assuming | |
| Apr 17, 2013 at 23:13 | comment | added | Martin Brandenburg | $F$ should also preserve the terminal object (=empty product). Just look up the definition of a group object in a category with products. | |
| Apr 17, 2013 at 11:02 | comment | added | user17945 | Not knowing much category theory, I'm afraid I don't understand this. Can you break it down for me. Are you saying if $F$ is a functor on the category of sets which satisfies $F(X\times Y) = F(X)\times F(Y)$, then $F(G)$ is automatically a group if $G$ is? If so, what is the identity in $F(G)$? | |
| Apr 17, 2013 at 6:32 | comment | added | Martin Brandenburg | This is trivial, functors preserving products preserve group objects. | |
| S Apr 17, 2013 at 3:04 | history | answered | user17945 | CC BY-SA 3.0 | |
| S Apr 17, 2013 at 3:04 | history | made wiki | Post Made Community Wiki by user17945 |