Timeline for Why pullback only defined up-to-isomorphism but nevertheless presented as functor?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Dec 11, 2012 at 18:07 | comment | added | Todd Trimble | Done. I also spelled out the universal argument at the end. | |
| Dec 11, 2012 at 18:05 | history | edited | Todd Trimble | CC BY-SA 3.0 |
added 652 characters in body
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| Dec 11, 2012 at 17:30 | comment | added | Almeo Maus | Yes thank you, I think it would be more clear for me with the same notations as in the question (at least at the beginning). Thanks for the reference!! I will read it as soon as tomorrow. | |
| Dec 11, 2012 at 17:17 | comment | added | Todd Trimble | If it helps, the universality argument at the end is a special case of Categories for the Working Mathematician (2nd ed.), p. 83, Theorem 2 (iv). It's good to go through the argument at least once in one's life. | |
| Dec 11, 2012 at 17:11 | comment | added | Todd Trimble | No, sorry, that was an accident in transcription. I can change it if you want (and hope there were no other errors in transcription). | |
| Dec 11, 2012 at 17:01 | comment | added | Almeo Maus | Thank you very much for your answer! However, I have to take some time to understand it. The $h$ in the question was $h : C' \rightarrow C$, but is it intentionally that you took $h : C \rightarrow C'$ ? | |
| Dec 11, 2012 at 16:24 | history | answered | Todd Trimble | CC BY-SA 3.0 |