Timeline for Can the standard map $\Sigma \Omega X \to X$ be a homotopy equivalence?
Current License: CC BY-SA 3.0
10 events
when toggle format | what | by | license | comment | |
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Sep 23, 2015 at 23:05 | comment | added | Greg Friedman | Thanks, that all makes sense - I'd just never seen that particular notation before. | |
Sep 23, 2015 at 16:03 | comment | added | Achim Krause | To add a third description (which is the one I use in my comment): If you embed $\mathbb{Z}/p^n\rightarrow \mathbb{Z}/p^{n+1}$, sending $1$ to $p$, you basically add in $1/p$ of the generator. $\mathbb{Z}/p^\infty$ is defined as the colimit over this sequence of maps, so it is filtered by cyclic subgroups of order $p^n$, each of which is $p$ times the next. That connects directly to both of Sean's perspectives. If you're familiar with how towers in the Adams spectral sequence encode cyclic groups of order $p^n$, the right intuition here is a tower that extends infinitely far down. | |
Sep 23, 2015 at 12:19 | comment | added | Sean Tilson | @GregFriedman One description is as all of the fractions where the denominator is a power of $p$ inside $\mathbb{Q}/\mathbb{Z}$. Equivalently, you can think of it as all of the $p^n$th roots of unity in $\mathbb{C}$ as you let $n$ vary. Does this help? | |
Sep 23, 2015 at 4:17 | comment | added | Greg Friedman | @AchimKrause: Maybe I should know this, but what exactly does the notation $\mathbb{Z}/p^\infty$ mean? | |
Sep 22, 2015 at 10:30 | vote | accept | jpaul | ||
Sep 22, 2015 at 10:12 | comment | added | Achim Krause | In general, a map that induces isos on homology with field cofficients induces isos on integral homology. A quick rundown: For $A\rightarrow B\rightarrow C$ a short exact sequence of abelian groups, if it works for two of these coefficients, it works for the third - that's the five-lemma. Start from $\mathbb{F}_p$ to inductively prove it for all $\mathbb{Z}/p^n$. Then it also follows for their colimit, $\mathbb{Z}/p^{\infty}$. But $\mathbb{Q}/\mathbb{Z}$ decomposes into copies of $\mathbb{Z}/p^{\infty}$, so we're done by $\mathbb{Z}\rightarrow \mathbb{Q}\rightarrow \mathbb{Q}/\mathbb{Z}$. | |
Sep 22, 2015 at 10:11 | comment | added | Fernando Muro | @AchimKrause OMG, I took an injective module for a flat one! | |
Sep 22, 2015 at 10:03 | comment | added | Achim Krause | The $Tor_1$ isn't zero: There's an exact sequence $Tor_1(\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) \rightarrow Tor_1(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \rightarrow \mathbb{Z}\otimes \mathbb{Q}/\mathbb{Z} \rightarrow \mathbb{Q}\otimes \mathbb{Q}/\mathbb{Z}$. The right- and leftmost terms are 0, so the Tor is just $\mathbb{Q}/\mathbb{Z}$. | |
Sep 22, 2015 at 9:52 | comment | added | Fernando Muro | There's a step I don't see. If you take $Y=M(\mathbb Q/\mathbb Z,n)$, $n\geq 2$, $Y^{\wedge 2}$ is contractible since $\mathbb Q/\mathbb Z\otimes \mathbb Q/\mathbb Z=0=Tor_1^{\mathbb Z}(\mathbb Q/\mathbb Z,\mathbb Q/\mathbb Z)$, isn't it? | |
Sep 22, 2015 at 9:35 | history | answered | Oscar Randal-Williams | CC BY-SA 3.0 |