Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If $X$ and $Y$ are based spaces, let $p_X: X\vee Y\to X$ be the fold map, or projection, onto $X$.

What is the homotopy fiber $F$ of $p_X$?

I think I have an argument that $F$ is the half-smash product $$Y\rtimes\Omega X := (Y\times\Omega X)/(\ast\times \Omega X).$$ But I'm not entirely sure of this and would like a reference, if one exists.

share|improve this question
3  
I was under the impression that the name "fold map" was reserved for $\mathrm{id}_X + \mathrm{id}_X : X\vee X \to X$. Am I mistaken? –  Ricardo Andrade Jun 27 '13 at 7:45
3  
I agree about the name "fold map", @RicardoAndrade, plus this map crushes $Y$ more than gently folding it. –  Omar Antolín-Camarena Jun 27 '13 at 11:54

4 Answers 4

I will consider overcategories of the form $Top/S$, $S\in Top$. Their objects are morphisms $R\to S$, their morphisms are triangles, commutative up to a (specified) homotopy, etc. For details see this. If we have a morphism $X \stackrel{f}{\to} Y$, then we have an adjoint pair of functors $$\begin{eqnarray*} f_* \colon Top/X & \leftrightarrows & Top/Y \colon f^*\\ f_* & \dashv & f^* \end{eqnarray*}$$

Here $f_*$ maps $Z\to X$ to the composition $Z \to X \to Y$, while $f^*$ maps $Z\to Y$ to $Z \times_Y X \to X$. Even more, we have a right adjoint $f^* \dashv f_!$, where $f_!$ is the "space of sections" functor. For $Y=\ast$ it maps $Z\to X$ to the space of its sections, and for general $Y$ it acts this way over every point $y\in Y$.

We don't need the exact definitions of these adjoints, we only need to know their existence, which implies that $f^*$ preserves all limits and colimits. Now notice that $Top\simeq Top/\ast$ and the fiber of $X\vee Y\stackrel{p}{\to} X$ is the same as $f^*(p)\in Top$, where $f\colon \ast \to X$ is the inclusion of point and we consider $p$ as an object of $Top/X$. We will now express $p$ as a colimit of simple objects and get the answer.

Notice that $X\vee Y = \mathrm{Colim}\left( X \leftarrow \ast \to Y \right)$ in $Top$, and the fold map is induced by maps $X = X$ and $Y\to \ast \stackrel{f}{\to} X$. I would draw a homotopy commutative diagram here, but I don't know how to do this on MO. Anyway, this is the same as the statement that the cofibered product of $X=X$, $\ast \stackrel{f}{\to} X$ and $Y\stackrel{\ast}{\to}X$ in overcategory $Top/X$ is $p$. Since $f^*$ preserves colimits, the fiber of $p$ is the same as the colimit of fibers of these maps. Now we have $$\begin{eqnarray*} f^*(X=X) & = & \ast \\ f^*(\ast\stackrel{f}{\to}X) & = & \Omega X \\ f^*(Y\stackrel{\ast}{\to} X) & = & Y \times \Omega X \end{eqnarray*}$$

Thus the fiber of $p$ is the same as the cofiber of embedding $\Omega X \to Y\times \Omega X$. This is the half-smash product you mentioned.

share|improve this answer

The synthetic summary of pretty much every approach described (this is just to give an easier forumalation of the actual calculation) is sometimes called Mather's Cube theorem, sometimes Ganea's (fiber) theorem, sometimes "flattening"; I like to call it distributivity, because it decategorifies to the equation $ x ( y + z) = (xy) + (xz)$.

Call the fiber you want $F$. The wedge sum $X\vee Y$ is a pushout \begin{array}{c} \bullet & \to & X \\ \downarrow & \ulcorner & \downarrow\\ Y & \to & X\vee Y \end{array} and the map $X\vee Y\to X$ (usually called "pinch") restricts as $1_X$ on $X$ and as $0: Y\to X$ on $Y$. By the Pasting Lemmas, the fibers of the composites $Y \to X$ and $ X \to X $ are the pullbacks of $F$ along $Y \to X\vee Y$ and along $X \to X\vee Y$, respectively, and of course the fiber of $ \bullet\to X$ is $\Omega X$, either way you factor it; but these fibers we know because they are easy to describe; by the universality of this last fiber, there is a coherent cube: \begin{array}{c} \Omega X & & & \to & & & \bullet \\ & \searrow & & & & \swarrow \\ & & \bullet & \to & X \\ \downarrow & &\downarrow & \ulcorner & \downarrow & & \downarrow \\ & & Y & \to & X\vee Y\\ & \nearrow & & & & \nwarrow \\ Y \Omega X & & & \to & & & F \end{array} What distributivity (flattening/Mather/Ganea) says is that, because the bottom square (small, inner square) is a pushout, and the vertical (distorted) squares are all pull-backs, and the whole cube is coherent, so the upper square (large) is, as the bottom square, a pushout. That is, $F\sim Y \Omega X / \Omega X $ is the half-smash you were suspecting.

It is important to make sure you understand the maps among the known fibers, when doing these calculations; not all maps $A B \to B$ are the obvious projection! In the present case, there just aren't too many maps $\Omega X \to \bullet$, and you can choose the structure of $\Omega X$ via the fiber-of $Y\Omega X \to Y$, as in the leftmost square.

As "some guy on the street", I once asked here how broadly does the distributive law hold, and Jacob Lurie's answer was that, according to Rezk, among homotopical categories those with the distributive law are probably homotopy toposes. I don't know how much that helps you, but searching for "Mather cube", or "Ganea fiber" or "flattening" will provide lots of relevant references.

share|improve this answer

I do not know of a reference but there is a very down to earth approach, which is to use the construction of the homotopy fiber of $f:S\to T$ as the subspace of $S \times PT$ given by $\{(s,\gamma)\mid \gamma : [0,1] \to T, \gamma(0)=\ast, \gamma(1)=f(s)\}$. For the fold map you get $Y\times \Omega X$ (for pairs $(s,\gamma)$ with $s\in Y$) and $PX$ (for pairs with $s \in X$) glued along $\Omega X$. Quotienting out by the contractible space $PX$, you get the thing you called the half smash product.

share|improve this answer
    
Thanks, this was exactly the argument I had in mind. Do you know a reference to a proof? –  Mark Grant Jun 27 '13 at 6:04
1  
No, I don't have a reference, and if it were me, I wouldn't think the time spent searching for one is really worth it: I'd just write a proof, either this one you had in mind or the the one Anton Fetisov and John Klein mentioned using that homotopy pushouts are stable under base change. To write up that proof I would just quote Mather's first theorem of the cube for the stability and then briefly explain as Anton did how this is a special case. –  Omar Antolín-Camarena Jun 27 '13 at 12:32

The easiest way I think to get what you want is to use Omar Antolín-Camarena's approach. I just want to point out that the result is actually a very special case of something that is much more generally true: in certain special cases, one can commute a homotopy limit with a homotopy colimit. (See below).

Suppose $I \to \text{Top}/X$ is a diagram of spaces over a fixed space $X$, where $I$ is a finite indexing category. Let the objects in the diagram be called $Y_\alpha$ where $\alpha$ varies through the objects of $I$. Let $$ Y = \text{hocolim}_{\alpha} Y_\alpha $$ be the homotopy colimit of the diagram. Then $Y \in \text{Top}/X$ is an object.

Then Suppose $X' \to X$ is a map. Define $Z_\alpha$ to be the homotopy limit $$ \text{holim}(X' \to X \leftarrow Y_{\alpha}) $$ This gives a diagram $I \to \text{Top}/X'$.

Assertion: The homotopy colimit of the diagram $\alpha \mapsto Z_\alpha$ coincides up to homotopy with the homotopy inverse limit of the diagram $$ X' \to X \leftarrow Y \, . $$ That is, homotopy colimits commute with base changes.

Your assertion follows from mine if we take $X'$ to be a point and $\alpha \mapsto Y_\alpha$ to be the diagram defining the wedge as a pushout.

As I recall, the assertion can be proved using a quasi-fibration argument. Another way is to use model categories.

Let's check a special case: suppose $I$ is the category $1 \leftarrow 12 \to 2$. The argument can be given in three steps:

Step 1: we can assume without loss in generality that $Y_1 \leftarrow Y_{12} \to Y_2$ is a diagram of Hurewicz fibrations over $X$ and that the arrows of the diagram are cofibrations. In particular, the homotopy colimit of this diagram coincides with its colimit.

Step 2: According to a theorem of Arne Strøm, the base change of a cofibration is a cofibration. This means that the diagram over $X'$ of base changes $$ Z_{1} \leftarrow Z_{12} \to Z_2 $$ consists of cofibrations, and each map to $X'$ is again a fibration. Again, the homotopy colimit of this displayed diagram coincides with the colimit.

Step 3: The colimit of the diagram displayed in Step 2 is clearly the base change of the colimit of the diagram displayed in Step 1.

share|improve this answer
1  
That homotopy pushouts are stable under homotopy pullback is one of Mather's theorem of the cube, and I think Mather's original proof of it is the proof using quasi-fibrations you remember. –  Omar Antolín-Camarena Jun 27 '13 at 11:52
    
@Omar: I'm not familiar with Mather's result. Can you give me a reference? At the end of the day, I believe result boils down to an old result of Dold and Thom on patching together quasifibrations. That certainly predates the work of Mather. –  John Klein Jun 27 '13 at 13:32
    
Got it: Pull-backs in homotopy theory. Canad. J. Math. 28 (1976), no. 2, 225–263 –  John Klein Jun 27 '13 at 13:35
    
Right, that's the paper I meant, @John. I haven't looked at it in a while, but as I recall Mather does quote the Dold and Thom results on quasi-fibrations. Also, I think most of the paper is the proof of the other, harder, Mather cube theorem. By the way, these cube theorems (well, the first one generalized to say all homotopy colimits are stable under base change, but the second one as is), pretty much characterize ∞-toposes. See this for a precise statement. –  Omar Antolín-Camarena Jun 27 '13 at 13:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.