Timeline for Homotopy type of some lattices with top and bottom removed
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jul 29, 2016 at 3:25 | comment | added | Omar Antolín-Camarena | (I'm sure Tom knows how to prove that claim he made, I just added the above proof for future readers, and this disclaimer did not fit into the same comment.) Oh, and $A$ and $B$ are contractible by Tom's remark that join with a contractible space is contractible, which also didn't fit into the previous comment. | |
| Jul 29, 2016 at 3:14 | comment | added | Omar Antolín-Camarena | To show that $C(P\times Q)\simeq\Sigma(C(P)\ast C(Q))$ first notice that the Grothendieck construction of the diagram of posets $X\leftarrow X\times Y\to Y$ is the poset $Z(X,Y):=(X\cup\{1\})\times(Y\cup\{1\})\setminus\{(1,1)\}$ where $1$ denotes a new top element to adjoin. So $Z(X,Y)$ has the homotopy type of $X \ast Y$. Then observe that $C(P\times Q)$ is the union of $A:=Z(C(P),C(Q)\cup\{0\})$ and $B:=Z(C(P)\cup\{0\}), C(Q))$ which intersect along $Z(C(P),C(Q))$ ($0$ is a new bottom element to adjoin). Since $A$ and $B$ are contractible, $C(P\times Q)\simeq\Sigma Z(C(P),C(Q))$. | |
| Jul 26, 2016 at 20:41 | comment | added | Tom Goodwillie | "Fresh points": what a great way to say it! | |
| Jul 26, 2016 at 13:36 | comment | added | მამუკა ჯიბლაძე | Oh I see you collapse them to fresh points, yes? | |
| Jul 26, 2016 at 13:34 | comment | added | მამუკა ჯიბლაძე | @GregoryArone I thought $\Sigma$ was the cylinder with collapsed lids. | |
| Jul 26, 2016 at 13:26 | comment | added | Gregory Arone | By the way, for the result to hold you have to assume that $P$ and $Q$ each have more than one element. You can include the one-element poset in the result by by decreeing that if $P$ has one element then $C(P)=S^{-2}$ is the $-2$-dimensional sphere. | |
| Jul 26, 2016 at 13:22 | comment | added | Gregory Arone | What's your favorite definition of $\Sigma X$? For example, you can defined it as the double mapping cylinder of the pushout diagram $*\leftarrow X \to *$. Now substitute $X=\emptyset$. | |
| Jul 26, 2016 at 13:12 | comment | added | მამუკა ჯიბლაძე | I became confused as soon as I accepted it :D How is $\Sigma(\varnothing)=S^0$? | |
| Jul 26, 2016 at 13:06 | vote | accept | მამუკა ჯიბლაძე | ||
| Jul 26, 2016 at 12:12 | comment | added | Gregory Arone | A reference to this result is Theorem 5.1 (d) in the paper by J. Walker, Canonical homeomorphisms of posets, European J. Combin. 9 (1988), no. 2. | |
| Jul 26, 2016 at 12:07 | history | answered | Tom Goodwillie | CC BY-SA 3.0 |