Timeline for Categories internal to schemes and subschemes of invertible arrows
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 20, 2017 at 9:03 | comment | added | David Roberts♦ | Thanks for coming back to this. With a bit more alg. geom. up my sleeves now, this is kind of obvious (lfp+formally étale composed with étale is clearly étale!) | |
| Jun 9, 2012 at 17:13 | comment | added | S. Carnahan♦ | Yes, if $s,t: X_1 \to X_0$ are étale, then so are the compositions with $X_1^{iso} \to X_1$. | |
| Jun 9, 2012 at 10:27 | comment | added | David Roberts♦ | For the previous comment, I mean in the case that the category $X$ has etale source and target maps. | |
| Jun 9, 2012 at 10:24 | vote | accept | David Roberts♦ | ||
| Jun 9, 2012 at 10:24 | comment | added | David Roberts♦ | Actually would I be correct in assuming that the groupoid $X^{iso}$ has etale source and target maps? This should follow from $X_1^{iso} \to X_1$ being formally etale, both maps $X_1 \to X_0$ being formally etale, and both maps $X_1^{iso}$ being lfp. | |
| Jun 9, 2012 at 9:05 | comment | added | David Roberts♦ | Hi Scott - I'm curious to know what happens if we additionally assume the source and target maps of the original category are etale... | |
| Jun 9, 2012 at 2:57 | comment | added | S. Carnahan♦ | Thanks for pointing that out. As it happens, my argument only needs the properties I have now enumerated. | |
| Jun 9, 2012 at 2:53 | history | edited | S. Carnahan♦ | CC BY-SA 3.0 |
removed bogus construction
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| Jun 9, 2012 at 0:08 | comment | added | David Roberts♦ | Hi Scott - your construction of $X_1^{iso}$ is not quite right. In particular, there is no swap map which is an endomorphism of $X_1 \times_{X_0} X_1$. Martin has it correct (although slightly overcomplicated - see my comments to his answer) when he uses $X_1 \times_{X_0\times X_0} X_1$ which is the pullback of $(s,t)$ and $(t,s)$, the source and target maps. One also needs to use the product $X_0 \times X_0 \to X_1 \times X_1$ of the unit map with itself. Does your argument still work? | |
| Jun 8, 2012 at 22:12 | history | answered | S. Carnahan♦ | CC BY-SA 3.0 |