Timeline for Is $S^1\vee S^1$ an Eilenberg-Mac Lane Space to a Homotopy Purist?
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9 events
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| Jul 13, 2016 at 23:46 | comment | added | Anton Fetisov | This is essentially the true reason. Considering the genuinely special $n=1$ thing in Tyler Lawson's comment in another answer, we similarly have that pointed $n-1$-connected spaces are equivalent to $E_n$-groups. Under this equivalence $K(A, n)$ is the discrete space $A$ with obvious $E_n$-group structure coming from abelian group structure. The equivalence tells us that $K(A,n)\vee K(B,n)$ is also $n$-connected and its $n$-fold loop space is the free product of $E_n$-groups $A$ and $B$. But a free product of $E_n$-groups is not discrete: already the free $E_n$-group $\Omega^n S^n$ isn't. | |
| Jul 12, 2016 at 18:01 | comment | added | Omar Antolín-Camarena | Well, sure @JesseC.McKeown: the homotopy fiber of the map $S^1 \vee S^1 \to BF_2$ has the homotopy type of the universal cover, so this proof is secretely about universal covers. What I meant is that even though you avoid having to mention covering space theory, the only way I figured out how to finish your argument is by building a model of the universal cover and showing it's a tree. So the meat seems the same and I was hoping maybe it wouldn't. | |
| Jul 12, 2016 at 14:09 | comment | added | Jesse C. McKeown | Well, effectively, yes, it would have to be the universal cover construction, in the end (what else is a contractible fibration over a graph?); but the point is we don't have to say "universal cover" or remember a recipe for building it. | |
| Jul 11, 2016 at 22:18 | comment | added | Omar Antolín-Camarena | I like your Ans 1! Although I have to say it felt very much like building the universal cover and observing it's a tree. What I mean is that the only way I saw to show the contractibility of the homotopy pushout of discrete spaces $F_2/\langle x\rangle \leftarrow F_2 \rightarrow F_2/\langle y \rangle$ is to prove that the usual double mapping cylinder construction of the homotopy pushout is a tree. | |
| Jul 11, 2016 at 21:52 | comment | added | Jesse C. McKeown | $A_\infty$ spaces can be strictified. So when both exist in the right category, the $A_\infty$-colimit is a competitor, in groups, to the group colimit, and the group colimit is a competitor, in $A_\infty$-spaces, to the $A_\infty$-colimit. Composing comparison maps on both sides et.c. | |
| Jul 11, 2016 at 21:48 | comment | added | Jesse C. McKeown | The spaces in the $A_\infty$-operad are contractible? ~~~~ Or, what happens if I say: let $G$ and $H$ be simplicial groups, and define $(G * H)_k = (G_k) * (H_k)$ with faces/degeneracies defined extending $G_k \to G_{k\pm 1} \to (G * H)_{k\pm 1} $ ... (which are therefore simplicial groups again)? ~~~~ Good questions, of course. Will get back to you. | |
| Jul 11, 2016 at 21:38 | comment | added | Qiaochu Yuan | Right, Omar's question is the reason this isn't a formal exercise (at least as far as I can see). Analogously to my answer, the inclusion of groups into grouplike $E_1$ spaces has a left adjoint given by $\pi_0$, so it preserves homotopy limits for formal reasons, but the question at hand is why it preserves a certain homotopy colimit. | |
| Jul 11, 2016 at 21:09 | comment | added | Omar Antolín-Camarena | For Ans 2, how do you show that the coproduct of $\mathbb{Z}$ and $\mathbb{Z}$ is $\mathbb{Z} * \mathbb{Z}$ in grouplike $A_\infty$-spaces? I know it's the coproduct in groups (or even monoids), can that statement be reduced to this one somehow? | |
| Jul 11, 2016 at 20:58 | history | answered | Jesse C. McKeown | CC BY-SA 3.0 |