Timeline for (Co)homological characterization of homotopy pullbacks
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Mar 11, 2012 at 9:23 | comment | added | Matthias Künzer | In K(Z-mod) (so not directly related to your question), Alberto Canonaco has found a commutative quadrangle with isomorphic cones that is not homotopy cartesian. Cf. Canonaco, K., "A sufficient criterion for homotopy cartesianess". | |
| Mar 2, 2012 at 22:01 | vote | accept | Mikhail Bondarko | ||
| Mar 1, 2012 at 20:27 | vote | accept | Mikhail Bondarko | ||
| Mar 1, 2012 at 20:27 | |||||
| Feb 28, 2012 at 23:24 | answer | added | John Klein | timeline score: 1 | |
| Feb 28, 2012 at 19:04 | answer | added | Ronnie Brown | timeline score: 1 | |
| Feb 28, 2012 at 15:58 | comment | added | Karol Szumiło | It's hard to say whether it addresses your question, but you may try to take a look at Mather Hurewicz Theorems for Pairs and Squares. | |
| Feb 28, 2012 at 14:48 | answer | added | Mark Grant | timeline score: 1 | |
| Feb 28, 2012 at 14:33 | comment | added | Mark Grant | Fixed diagram using array (pro-tip: you have to use \newline instead of \\ to move to the next row). | |
| Feb 28, 2012 at 14:32 | history | edited | Mark Grant | CC BY-SA 3.0 |
fixed diagram
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| Feb 28, 2012 at 13:51 | answer | added | Tyler Lawson | timeline score: 4 | |
| Feb 28, 2012 at 13:38 | comment | added | Mikhail Bondarko | This seems to be too complicated. Actually, I just mean a very basic commutative square, with arrows from $A$ to $B and $C$ and from $B$ and $C$ to $D$. | |
| Feb 28, 2012 at 13:29 | comment | added | Mark Grant | I can't read your comment. Some hints on how to include diagrams can be found here: tea.mathoverflow.net/discussion/871/… (I should say I haven't tried this yet myself). | |
| Feb 28, 2012 at 13:28 | comment | added | Federico Poloni | You need to include an image --- I am afraid that the TeX interpreter on the forum cannot parse commutative diagrams. | |
| Feb 28, 2012 at 13:14 | comment | added | Mikhail Bondarko | I am sorry; I would like to write $$S= \begin{CD} A@>{}>> B \\ @VV{}V @VV{}V \\ C@>{}>>D \end{CD}$$ but this doesn' seem to work. | |
| Feb 28, 2012 at 13:05 | comment | added | Mark Grant | I don't quite see how your "commutative square" is a square. Could you explain your notation, please? | |
| Feb 28, 2012 at 12:47 | history | asked | Mikhail Bondarko | CC BY-SA 3.0 |