Timeline for Probably easy: Why is f*:A^C'->A^C continuous and cocontinuous for any functor f:C->C'?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Jul 24, 2010 at 10:12 | comment | added | Harry Gindi | Yeah, you flipped the variance. | |
| Jul 24, 2010 at 10:04 | comment | added | Martin Brandenburg | I just thought about the example $f_* : Sh(X) \to Sh(Y)$ for a continunous map $f : X \to Y$. | |
| Jul 24, 2010 at 8:56 | history | made wiki | Post Made Community Wiki by Harry Gindi | ||
| Jul 24, 2010 at 8:53 | vote | accept | Harry Gindi | ||
| Jul 24, 2010 at 8:52 | answer | added | Peter LeFanu Lumsdaine | timeline score: 5 | |
| Jul 24, 2010 at 7:58 | history | edited | Harry Gindi | CC BY-SA 2.5 |
edited title
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| Jul 24, 2010 at 7:57 | comment | added | Harry Gindi | I don't think that's right. f*=Hom(f,A). It has an upper star because it is the image under a contravariant functor. This notation is standard in descent theory as well (cf. Stacks-GIT, for instance). | |
| Jul 24, 2010 at 7:50 | comment | added | Martin Brandenburg | I think your $f^*$ should be denoted $f_*$. Then it's left adjoint is given by a Kan extension. | |
| Jul 24, 2010 at 7:44 | history | asked | Harry Gindi | CC BY-SA 2.5 |