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Given a map of $n$-fold loop spaces $X\to Y$, we can take the homotopy cofiber, denote it $Y/X$ (all spaces here will also have a base point, and all maps pointed). I have some basic questions about this construction:

  1. When is it the case that $X/Y$ is again an $n$-fold loop space?
  2. Are there weaker conditions that guarantee $X/Y$ is an $m$-fold loop space for some $0<m<n$?
  3. As a topological space, $X/Y$ satisfies a universal condition up to homotopy. In particular, it induces a morphism $X/Y\to Z$ whenever there is a morphism $X\to Z$ such that the composition $Y\to X\to Z$ is null. When is it true that this induced morphism is again a morphism of $n$- or $m$-fold loop spaces, assuming $X/Y$ is?

I feel like these should be some kind of "homotopy normality" conditions on $Y\subset X$, which I know that others have studied (e.g. Presma) for $n=1$.


(EDIT)

So, I don't want to delete the above, because I think it generated some useful stuff even thought it has some obvious mistakes and confusion in it.

But I want to clarify and try to correct my question. My question really is, I guess, the following: given a closed inclusion of Lie groups $H\to G$, the coset space $G/H$ is also the base space in a fibration $H\to G\to G/H$ (really a principal $H$-bundle). So in this situation, we have this sequence of spaces $H\to G\to G/H$ which satisfies (it seems to me) two universal properties. On one hand, it's a fibration, so in other words, $H$ is, up to homotopy, the fiber over the projection map $G\to G/H$. On the other hand, $G/H$ is precisely the coset space, so given a group homomorphism $\phi:G\to F$ such that $H$ goes to $1_F$, it would seem that one should obtain a factorization $G/H\to F$. Is this not true? In particular, though $G/H$ is not the homotopy cofiber of the map $H\to G$, it is $G/\sim$ for some equivalence relation, thus it admits such a universal property. More generally any continuous map $G\to Z$ such that any two points which differ by an element of $H$ go to the same place must factor through $G/H$ it seems.

Now really my questions is: how much of this relies on the property of being Lie groups? Can't we just do this for $E_n$-algebras in spaces? In general, is there some kind of homotopical Borel construction for a map of $E_n$-algebras which factors certain maps out of the "total space"?

And moreover, what conditions do we need on the inclusion of the sub-$E_n$-algebra to make sure that the quotient is still $E_n$ or $E_m$ for $m<n$?

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  • $\begingroup$ What if the map $f:X \to Y$ is a m-fold loop map, i.e. $f = \Omega^m g$ for some $g$ and $m\leq n$? Isn't then $Y/X$ a $m$-fold loop space? $\endgroup$
    – Prasit
    Sep 9, 2015 at 4:03
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    $\begingroup$ @Prasit: no. For example, take $f : X \to Y$ to be an inclusion of discrete groups (so $n = 1$). The homotopy cofiber of $f$ is $Y$ with the image of $X$ identified to a point, which is very different from the group-theoretic quotient $Y/X$ (which still need not be a group of $X$ isn't normal). That is, there's a forgetful functor from $n$-fold loop spaces to spaces, and it's not even close to preserving homotopy cofibers. $\endgroup$ Sep 9, 2015 at 4:09
  • $\begingroup$ When you say "$n$-fold loop space", do you mean that literally or only up to homotopy equivalence? I would have assumed the latter, except that if so I don't know why you tagged the question "general topology"... $\endgroup$ Sep 9, 2015 at 4:33
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    $\begingroup$ Another example is $S^1 \to S^1$ via the double covering map. $S^1 = K(\mathbb{Z},1)$ is an infinite loop space and this is an infinite loop map, but the cofiber is $\mathbb{RP}^2$. I don't think that is even a 1-fold loop space. Do you have any examples of this happening where $n>1$, $m>0$, and the quotient map $Y \to Y/X$ is an m-fold loop map? (presumably you meant for this last condition in your question, otherwise $Y/X$ could just accidentally happen to have an m-fold loop space structure with nothing to do with $X$ or $Y$). $\endgroup$ Sep 9, 2015 at 7:18
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    $\begingroup$ Admittedly, I now see that the homotopy cofiber is the wrong thing to be doing here. $\endgroup$ Sep 9, 2015 at 19:25

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I think what Jesper and Amrani wrote should clarify why taking the cofiber is not very appropriate in this context and what Qiaochu wrote clarifies why taking the fiber of the delooping is "better".

Let me try to complement the picture. I apologize in advance for the long answer, but since we already talked about it, I feel I may not have made my self clear enough.

The recognition principle tells us that the homotopy theory (say, relative category) of $n$-loop spaces is equivalent to that of pointed $(n-1)$-connected Kan complexes. When $n=1$, the functor in one direction is $\Omega$ and we can denote its inverse by $B$ as Qiaochu did.

A more rigid equivalence between these theories (say for $n=1$) is given by the Kan loop group $\mathbb{G}$ and $\bar{W}$; for $n>1$, you essentially apply these functors successively. What this means is that you can ridigify every $n$-fold loop space into a simplicial group in a functorial way. Of course, playing with sing and $|-|$ allows you to get even a topological group model. There is some care to take here though -- rigidification is a \textbf{model dependent} question: if you take ordinary group objects in the homotopy theory of spaces as presented by $Cat$ and Thomason equivalences (and the induced homotopy theory), you will not get something equivalent to group-like $\mathbb{E}_1$-algebras but rather the homotopy theory of crossed modules aka pointed connected $2$-types.

So let's say you're in $Top$ or $sSet$. What is a principle fibration structure on a map $E \to B$ of pointed connected spaces? well, up to (coherent) homotopy, it is the same as a group object $G$ and a map $B\to BG$ such that $E\to B\to BG$ is a homotopy fiber sequence in a prescribed way. To say this "model independently", you should say that it is a pointed map $B\to X$ for some pointed connected $X$ with the structure described above and call such a map an h-principle fibration (Nikolaus, Schreiber and Stevenson call it principle $\infty$-bundle).

Now let us come back to your question. Suppose you start with a map of $n$-fold loop spaces $\Omega^n f:\Omega^n E\to \Omega^n B$ (that is $E\to B$ is a map of pointed $(n-1)$-connected spaces). The homotopy quotient $\Omega^n B//\Omega^n E$, is then taken to be $hofib(\Omega^{n-1}E \to \Omega^{n-1} B)$ and as such carries a structure of an $(n-1)$ fold loop space. Note that if $E\to B$ was an h-principle fibration with structure $B\to X$, then $\Omega^n B//\Omega^n E\simeq \Omega^n X$ and the map $\Omega^n B\to \Omega^n B//\Omega^n E$ would be equivalent to an $n$-fold loop map by your initial prescribed structure.

You rightfully ask: how can we view $\Omega^n B//\Omega^n E$ as a (homotopy) Borel construction? Well, if we're in one of the "rigidifyable" homotopy theories of spaces, we could first convert $\Omega^n E\to \Omega^n B$ into a map of group objects $H \to G$ and then take the bar construction $Bar(H,G)=|Bar_\bullet(H,G)|$ which is evidently a model for the Borel construction $G\times_H EH$. But how do you do it without rigidifying? well, assuming we defined the homotopy quotient as the fiber, we can take the Cech nerve $C_\bullet^h(\Omega^n B\to \Omega^n B//\Omega^n E )$ (here I really mean the homotopy coherent version, e.g. apply the ordinary Cech nerve after you replaced the map to a fibration). A perhaps surprising thing is that as a simplicial space it is weakly equivalent to $Bar_\bullet(H,G)$ (i.e. there is a zig-zag of simplicial maps) -- this means in particular that in simplicial degree $k$ it is equivalent to a product $(\Omega^n E)^k \times \Omega^n B$ and not just a fiber product. So it is natural to call $C_\bullet^h(\Omega^n B\to \Omega^n B//\Omega^n E )$ the homotopy action groupoid and refer to its realization as a Borel construction of the initial map $\Omega^n E\to \Omega^n B$. This gives you the universal property you alluded to.

This last trick works more generally: if you start with an h-principle fibration $E\to B$ and you rigidify it (say in simplicial sets) to a principle fibration $G\to E'\to B'$ then the Cech nerve $C_\bullet^h(E\to B)$ would be equivalent to the bar construction $Bar_\bullet(G, E')$. (Edit) Note that this means that for every h-principal fibration $E\to B$, $C_\bullet^h(E\to B)$ is a Segal space (like the ordinary bar construction of a group action is). But there is an other Segal space around -- $C_\bullet^h(*\to X)$ -- which is in fact a group-like reduced Segal space (it is a "Segal group" that presents all the coherent multiplication of $\Omega X$). If we assume (for example) $B\to X$ was a fibration of Kan complexes with fiber $E$ then we get a map of Segal spaces $C_\bullet^h(E\to B)\to C_\bullet^h(*\to X)$. So we are lead to some kind of notion of a Segal space over a Segal group. The homotopy theory of these would be equivalent to that of spaces over the realization of the Segal group and thus gives you a model for coherent group actions. What's happening in the setting of a loop map $\Omega f:\Omega E\to \Omega B$ is that the loop space $\Omega E$ coherently acts on the space $\Omega B$ via the map $\Omega f$ (just like the domain of a group map acts on the underlying set of the codomain). The homotopy quotient of this coherent action is what we denoted $\Omega B//\Omega E$.

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  • $\begingroup$ Brilliant! Thanks so much Matan! :) $\endgroup$ Sep 10, 2015 at 23:23
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    $\begingroup$ Thanks! I elaborated a bit on the role of the Cech nerve. I think it's getting close to a shameless self advertisement, so I'll stop here ;) $\endgroup$
    – MatanP
    Sep 11, 2015 at 10:58
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Even for $n=1$ the cofiber will almost never be a loop space.

There is no reason for it to be a loop space, and plenty of reasons for it not to be a loop space. E.g., it would need to be a simple space, its cohomology would need to be a Hopf algebra,.... These are usually not satisfied in examples; see the comments, eg. Chris' example $RP^2$.

What Presma studies for $n=1$ is the different question of when $f: X \to Y$ is part of a fibration sequence $X \to Y \to Z$ not a cofibration sequence. (These only agree in the stable range.) Is this what you are really after? I agree that there should be a higher version of that theory.

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  • $\begingroup$ Hm, I'm sort of after when such things coincide. For instance, when one has an inclusion of Lie groups $H\to G$, and the space of cosets is also a Lie group. This seems to be not particularly uncommon. $\endgroup$ Sep 9, 2015 at 13:21
  • $\begingroup$ In other words, sometimes fibrations look like group quotients. $\endgroup$ Sep 9, 2015 at 13:30
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    $\begingroup$ @JonBeardsley, the space of cosets G/H is not the homotopy cofiber, so they do not coincide in that case. The homotopy cofiber just cones off the subspace H of G. $\endgroup$ Sep 9, 2015 at 14:49
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May be I going to say something obvious, but I hope it will give some clarification since I feel (Maybe I'm wrong) that there is some confusion about where you are taking the homotopy cofiber regarding your comments about Lie groups.

If you have a map of loop spaces (i.e. an $A_{\infty}$-map) say $f: \Omega X\rightarrow \Omega Y$ (where $X$ and $Y$ are connected pointed spaces)

The homotopy cofiber in the $\infty$-category of $A_{\infty}$-spaces (group like) is computed as follows. First you take the bar construction and you obtain $Bf: X\rightarrow Y$, then you compute the homotopy cofiber in the category of pointed topological spaces which I will denote by $Y/X$. Finally, the homotopy cofiber (quotient) of $f: \Omega X\rightarrow \Omega Y$ in $\infty$-category of $A_{\infty}$-spaces is $\Omega (Y/X)$.

The fiber sequence $\Omega SU(n)\rightarrow \Omega SU(n+1)\rightarrow \Omega S^{2n+1} $ can not be seen as a quotient group in any reasonable sense. A reasonable notion of quotient group of $\Omega SU(n)\rightarrow \Omega SU(n+1)$ in the category of $A_{\infty}$-spaces is $\Omega (SU(n+1)/SU(n))$ (where $SU(n+1)/SU(n)$ is the quotient in the category of pointed spaces so it is not the space of cosets.fee)

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  • $\begingroup$ It can't be thought of as a quotient in any reasonable sense? Is it at least true that homogeneous spaces coming from fibrations can be thought of as quotients? $\endgroup$ Sep 9, 2015 at 18:56
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    $\begingroup$ @JonBeardsley It all depends on what you mean by "reasonable" and what you mean by "quotient" ;) What is true is that if F -> E -> B is a principal fibration (like the sequence above) then F is equivalent to group which acts on E, and B can be identified with the Borel construction of the action of F on E. But what Amrani is saying is that this is not taking a cofiber, neither in spaces nor A_infty space. $\endgroup$ Sep 9, 2015 at 19:17
  • $\begingroup$ @JesperGrodal okay, good. At least I'm not totally lost here... so I suppose my question is that a lot of times even the group theoretic quotient has such a universal property, and I'm wondering if it has any such thing here. $\endgroup$ Sep 9, 2015 at 19:25
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One way to get $G/H$ from the map $f : H \to G$ is to first deloop it, getting $Bf : BH \to BG$, and then take homotopy fibers, getting a fiber sequence

$$H \to G \to G/H \to BH \to BG.$$

This suggests the correct thing to do for (grouplike) $E_n$ spaces, which is the following: if $f : X \to Y$ is an $E_k$ map between grouplike $E_n$ spaces, $k \le n$, then delooping $k$ times gives us a map $B^k f : B^k X \to B^k Y$, and taking homotopy fibers gives a fiber sequence

$$\dots \to B^{k-1} X \to B^{k-1} Y \to Z \to B^k X \to B^k Y.$$

It follows that $\Omega^{k-1} Z$, the homotopy fiber of the map $Bf : BX \to BY$, is by construction naturally $E_{k-1}$. This is a reasonable candidate for what one might call the "$E_1$ quotient" of $Y$ by $X$, but it loses information compared to, say, the information of $Z$ itself. One might call $\Omega^{k-i} Z$ the "$E_i$ quotient," and as $n, k \to \infty$ all of these quotients can themselves be packaged into the data of an infinite loop space.

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  • $\begingroup$ This is really nice @Qiaochu, thanks! $\endgroup$ Sep 10, 2015 at 18:10
  • $\begingroup$ Does this construction admit the universal property that we might expect of a quotient? $\endgroup$ Sep 10, 2015 at 18:34
  • $\begingroup$ @Jon: I think so; it should end up being the homotopy quotient of $G$ by the multiplication action of $H$. $\endgroup$ Sep 10, 2015 at 19:43

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