Timeline for Product of two group morphisms not a group morphism
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Apr 12, 2019 at 2:03 | vote | accept | Almeo Maus | ||
| Apr 11, 2019 at 16:27 | history | edited | Arturo Magidin | CC BY-SA 4.0 |
"Mac Lane", not "MacLane"
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| Apr 11, 2019 at 15:16 | answer | added | YCor | timeline score: 0 | |
| Apr 2, 2019 at 0:04 | comment | added | Almeo Maus | @MikeShulman Thank you so much, this is the answer I was looking for! Please make it an official one so that I can mark the question as answered. | |
| Apr 1, 2019 at 10:00 | review | Close votes | |||
| Apr 12, 2019 at 3:05 | |||||
| Apr 1, 2019 at 9:42 | comment | added | Mike Shulman | The cones are in $\rm Set$, but the set of cones, under pointwise multiplication is the limit in $\rm Grp$. In fancier language, he's saying that limits in $\rm Grp$ are created in $\rm Set$. | |
| Apr 1, 2019 at 8:07 | comment | added | Almeo Maus | And I mixed up because the composition is the reverse of the writing order, but the idea is there: in this case $f(1)^{-1}g(1)^{-1}f(1)g(1) = trans(\sqrt{2}*e_x+\sqrt{2}*e_y) \circ rot([0,0], \pi/4)$ | |
| Apr 1, 2019 at 8:01 | comment | added | Almeo Maus | @YCor Yes indeed, if the source group is $(R,+)$, if the destination group is the direct isometries of the $R^2$ plane, $E^+(2)$, then if we take $f(1)=rot([0,0],\pi/2)$ and $g(1)=trans(1*e_x)$, we get $g(1)f(1)g(1)^{-1}f(1)^{-1}\neq id$ | |
| Apr 1, 2019 at 7:32 | comment | added | YCor | "There's no reason why $x\mapsto g(x)f(x)$ should be a group homomorphism": you should easily find examples to confirm this expectation. | |
| Apr 1, 2019 at 7:30 | history | edited | YCor |
edited tags
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| Apr 1, 2019 at 7:26 | history | edited | Almeo Maus | CC BY-SA 4.0 |
added 348 characters in body
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| Apr 1, 2019 at 7:02 | comment | added | Almeo Maus | Ok, thank you. In this case, this is what lost me. I still don't see the point to be inside Sets, as we already know that Sets have all small limits. It would have been of much value to construct the limit in Grp, which would have been an example of how to do it without the use of the one-point set. | |
| Apr 1, 2019 at 6:57 | comment | added | Friedrich Knop | MacLane writes "...the set L of all cones (all matching strings x)...". So, he seems to mean cones in the category of sets, not groups. | |
| Apr 1, 2019 at 6:51 | history | edited | Almeo Maus | CC BY-SA 4.0 |
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| Apr 1, 2019 at 6:42 | history | asked | Almeo Maus | CC BY-SA 4.0 |