Timeline for Does Spec functor sends pushouts of rings into pullbacks of sets?
Current License: CC BY-SA 4.0
13 events
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Jan 12, 2020 at 11:29 | history | edited | Fabio Lucchini | CC BY-SA 4.0 |
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Jan 12, 2020 at 11:26 | vote | accept | Fabio Lucchini | ||
Jan 10, 2020 at 22:29 | history | edited | Fabio Lucchini | CC BY-SA 4.0 |
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Jan 10, 2020 at 22:21 | history | edited | Fabio Lucchini | CC BY-SA 4.0 |
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Jan 10, 2020 at 22:13 | history | edited | Fabio Lucchini | CC BY-SA 4.0 |
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Jan 10, 2020 at 22:08 | answer | added | Martin Brandenburg | timeline score: 3 | |
Jan 10, 2020 at 22:06 | comment | added | Fabio Lucchini | @MartinBrandenburg: Here $X$ is a generic set (or topological space); I'm trying to prove that the square of spectrum has the universal property of a pullback square in the category of sets (or topological spaces). | |
Jan 10, 2020 at 22:03 | comment | added | Martin Brandenburg | What is $X$ in your diagram? | |
Sep 22, 2019 at 9:30 | history | edited | Fabio Lucchini | CC BY-SA 4.0 |
improved clarness
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Sep 22, 2019 at 8:05 | comment | added | Gro-Tsen | So maybe your real question should be: under what conditions on $A\to C$ is it true that Spec takes $B\otimes_A C$ to the fiber product of topological spaces for all $A\to B$. But whatever the case, you should clarify your question because, so far, it's not clear what you're asking. | |
Sep 22, 2019 at 8:03 | comment | added | Gro-Tsen | If the second question is your real question, you can't seriously hope to do a characterization with $A\to C$ only, since the situation is symmetric and $A\to B$ plays exactly the same rôle! If $A\to B$ satisfies one of the sufficient conditions you listed and $A\to C$ does not, then the diagram will still be a pullback of topological spaces! | |
Sep 22, 2019 at 8:00 | comment | added | Gro-Tsen | Your question is unclear: first you seem to ask the general question (viz.: is it always true that Spec sends tensor products to pullbacks of topological spaces?), to which the answer is a resounding “NO” ($\mathbb{A}^2_{\mathbb{C}}$ as a topological space is not the product of two copies of $\mathbb{A}^1_{\mathbb{C}}$), then you ask about a specific equivalence: so, which is your question? | |
Sep 22, 2019 at 7:24 | history | asked | Fabio Lucchini | CC BY-SA 4.0 |