Timeline for Is there a functorial proof that Eilenberg-MacLane spaces are unique up to homotopy equivalence?
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6 events
| when toggle format | what | by | license | comment | |
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| Dec 11, 2010 at 5:15 | comment | added | Akhil Mathew | (Harry Gindi's comment to the question takes care of the case of simplicial abelian groups, if I understand correctly.) | |
| Dec 11, 2010 at 5:12 | comment | added | Akhil Mathew | Out of curiosity, is it any easier to prove the corresponding result for simplicial sets? | |
| Dec 10, 2010 at 19:55 | comment | added | André Henriques | Sorry. Scratch what I said above. The homotopy groups of $Map(H(A,n),H(A,n))$ are all in negative degrees. So the space of (pointed) maps between two homotopy equivalent Eilenberg-MacLane spaces $K(A,n)$ is actually contractible. | |
| Dec 10, 2010 at 19:47 | comment | added | André Henriques | It is NOT true that any other model is canonically homotopy equivalent to this one. Indeed, unless $n=1$, the connected components of the space of (pointed) maps between two homotopy equivalent Eilenberg-MacLane spaces $K(A,n)$ is not contractible. If you work at the level of the homotopy category, i.e., identify homotopic maps, then, of course, the problem goes away. | |
| Dec 9, 2010 at 22:34 | comment | added | David Roberts♦ | Does this help show why any other model of a K(A,n) is homotopic to the canonical one? I imagine it might, but I'm being a bit thick at the moment (see also my comment on the question) | |
| Dec 9, 2010 at 22:24 | history | answered | André Henriques | CC BY-SA 2.5 |