@PaulVanKoughnett so let Y_n be the quotient of [0,1] by the subspace {0, 1, 1/2, ... 1/n}; it's a bouquet of n circles. let X be the colimit: it is the quotient of [0,1] by the whole set Z = {0,1,1/2,1/3,...} (and happens to be homeomorphic to the Hawaiian earring). then the identity X -> X doesn't lift to a map X -> Y_n for any n. the maps Y_n -> X are bijective away from the open, dense, complement of Z, and so the only potential continuous lift is induced by the identity map of [0,1].
if you only mean homotopy equivalence, then it's still true but you have to work a little harder (all the Y_n are semilocally simply connected & so there's no homotopy-section X -> Y_n because it couldn't be injective on pi_1)
@CharlesRezk Btw I just had to work out how to cite this old chat message, because I prove in my thesis that the answer is yes: for any compact Lie group $G$, genuine $G$-spectra are the free stable compactly-generated symmetric monoidal $\infty$-category with duals for compact objects on genuine $G$-spaces (considered as a compactly-generated symmetric monoidal $\infty$-category)
Hello, I recently posted a question related to stable homotopy of classifying spaces. If somebody is interested, please see mathoverflow.net/questions/345599/…
@TimCampion Cool! And thanks for linking back to that discussion. I had completely forgotten that I had had such thoughts at one time :D
Is there a way to download the logs of this chatroom? I need to go collect all the places where I think stuff out loud in here, so they are not lost if this chatroom disappears someday.
It seems to me that, given an $\infty$-category $\mathcal{C}$ and two of its objects $A,Y\in \mathcal{C}$, the $\infty$-category $\mathcal{C}_{A//Y}$ whose $n$-simplices are maps of the form $\Delta^0\star \Delta^n \star \Delta^0 \cong \Delta^{n+2}\to \mathcal{C}$ in $\mathcal{C}$ with value $A$ at $0$ and $Y$ at $n+2$ is equivalent to the pullback $\infty$-category $\mathcal{C}_{A/}\times_{\mathcal{C}} \mathcal{C}_{/Y}$.
I believe I have a proof, but it also appear to me that I can derive some inconsistencies from this equivalence, so I am not so sure anymore
If that is an equivalence then, given maps $f\colon A\to B , g\colon A \to X, q\colon B \to Y$ and $p\colon X \to Y$ in $\mathcal{C}$ one would get $\mathcal{C}_{A//Y}((q,f),(p,g))\simeq \mathcal{C}_{A/}(f,g)\times_{\mathcal{C}(B,X)}\mathcal{C}_{/Y}(q,p)$, right?
So I wanna show that $f\perp p$ (defined by asking the square induced on hom-spaces by pre-composition with $f$ and post-composition with $p$ to be a pullback) if and only if $\mathcal{C}_{A//Y}((q,f),(p,g))\simeq \ast$ for every commutative square $(g,q)\colon f \to p$.
This does not seem to hold true, unless I made some mistake somewhere. Still, it feels like the definition in terms of those mapping spaces should be correct (I know Lurie does something similar, but the lack of details has lead me to try and prove this).
sorry let me just parse for a second (you can keep writing thought)
**though
ok this looks fine to me, I still don't see a contradiction (I guess it's worth pointing out I'm taking homotopy fibers everywhere, if that matters)
i.e. presumably you want "$f\perp p$" when the relevant square is homotopy cartesian, and also the formula for the mapping space in $\mathcal{C}_{A//Y}$ is a homotopy pullback above (though maybe it's an honest pullback depending on which of the many equivalent models for hom space you use? dunno)
I wanna show, for example, that if for every commutative square as above we have contractibility of that mapping space in $\mathcal{C}_{A//Y}$, then postcomposition by $p$ induces an equivalence $\mathcal{C}_{A/}(g,f)\simeq \mathcal{C}_{A/}(g,pf)$ for every $f\colon A\to X$.
I end up with some obstruction on the map induced on homotopy fibers, which (weird) depends on the automorphisms of the diagonal $pf$ of the square.
I don't think I'm getting any weird dependence on the automorphisms of the diagonal, but I guess I could've missed something tracing through all the different identifications. Maybe you, at some point, dropped the part of the description of the homotopy fiber as including the data of a specified equivalence?
In the vein of "why aren't there any weird loops in this homotopy limit", I had to trace through a few times to convince myself that for an object $X$ in a semiadditive $\infty$-category, the homotopy equalizer of $X \oplus X \rightrightarrows X$, where one arrow is the codiagonal and the other is the zero map, yields the cofree group object on $X$.
If M is a cancellative monoid (in sets) and N is a submonoid, then the canonical map from the homotopy orbits to the quotient M//N -> M/N is an equivalence (I think). Is there a slick way of seeing this? Ideally without considering the embedding into the group completions...
Is it true though? If x+a = y+b with a,b in N, does there exist z such that x in z+N and y in z+N? Without being able to subtract this is not so clear to me.
Hm but that's not what the quotient actually means ... monoids are confusing.
My idea was that if I look at the group completions, then the result is true. This implies that the bar construction has an extra degeneracy. But that degeneracy "cannot introduce minus signs", because of the form of the interaction of the degeneracy maps and boundary maps.
Just need to apply a filter that turns it into .tex, then make a .pdf, then put it on the arXiv, so everyone's brilliant insights will be preserved forevere
Suppose a discrete monoid $M$ acts "freely" on a set $X$, by which I mean that $mx=m'x$ for some $x$ implies $m=m'$. I think I want to show that $X_{hM}\approx X/M$, i.e., that $X_{hM}$ is discrete.
We have two simplicial objects: $A_n= M^n\times X$ and $B_n=X^{n+1}$, with a map $A\ra B$, all defined in the "obvious" way, e.g., M\times M\times X\to X\times X\times X sends $(m_2,m_1,x)\mapsto (m_2m_1x,m_1x,x)$.
Hold on a student is here ...
... a map $A\to B$ defined the "obvious" way.
No this is dumb.
Just think about $A_n \to X/M$. The fiber over any $[x]\in X/M$ is a subobject $S\subseteq A$, with $S_n=\{(m_n,\dots,m_1,y)\;|\; m_1,\dots,m_n\in M,\; [y]=[x]\}$.
You want to show each $|S|$ is contractible for each $[x]\in X/M$.
So $S$ is clearly the nerve of a category. Since $M$ acts freely on $X$, it is actually a poset.
Suppose for any $m_1,m_2\in M$, there exist $m_2',m_1'$ such that $m_2'm_1=m_1'm_2$. Then $S$ is a directed poset, so $|S|\approx *$.
@TylerLawson @TomBachmann Let R=Z[t^2, t^3]. I think there is a periodic resolution of Z[t] over R like Z[t]<--R^2<--R^2<--R^2<--... Now tensor over R with Z, being careful to remember that t^2 and t^3 get mapped to 1 inside Z, not to 0, and I believe the sequence is exact except at the beginning where you compute the quotient (like, you just get the matrix [1&1//-1&-1] everywhere, which is a projection, whence the exactness)
