Timeline for Categories internal to schemes and subschemes of invertible arrows
Current License: CC BY-SA 3.0
2 events
| when toggle format | what | by | license | comment | |
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| Jun 9, 2012 at 0:10 | comment | added | David Roberts♦ | Hi Martin, you can simplify your cartesian square so that the right hand vertical map is just $X_0\times X_0 \to X_1 \times X_1$ if you replace $\alpha$ by $\alpha'(f,g) = (m(f,g),m(g,f))$. But perhaps this doesn't help for your proof strategy. | |
| Jun 8, 2012 at 13:57 | history | answered | Martin Brandenburg | CC BY-SA 3.0 |