# (Homotopy)-Fibre of H-spaces

By H-space I mean a unital magma in $Ho(\mathsf{Top}_*)$, i.e. the homotopy category of pointed spaces (feel free to use any of your favourite model :) ).

This means that we have a pointed space $X$ and a map $\mu_X:X \times X \to X$ such that $\mu_X \circ j \simeq \nabla$, where $\nabla : X \vee X \to X$ is the fold (or co-diagonal) map, and $j:X \vee X \to X \times X$ is the pushout-product of the map $* \to X$ with itself.

An H-map between H-spaces is a map $f:X \to Y$ of the underlying spaces such that the obvious square involving $f$ and the multiplication maps is homotopy commutative.

Here comes the question: how to prove that the homotopy fibre of $f$ is again an H-space (in a sensible way, of course)?

Homotopy pullbacks of spaces are weak pullbacks in the homotopy category, and that's how you get a multiplication on the fibre compatible with that of $X$. However, I can't seem to find how to check the desired (homotopy) commutativity, as that would seem to require that weak pullback to be a strict one, which is known to be false.

• Do you know this to be true? Jan 26, 2017 at 11:23
• It's claimed in lots of places (without proof), though I am starting to question its validity Jan 26, 2017 at 11:26
• Is there something wrong with this argument? Replace the map f with a homotopy equivalent fibration. This should still be a map of H-spaces, and now the actual fiber is the homotopy fiber and also an H-space. It's homotopy equivalent to the homotopy fiber of the original map, whence the claim. Jan 26, 2017 at 16:31
• The fact that the fibre of the replaced map is an H-space is exactly what I am trying to prove, but I can't get unitality under my assumptions (I basically get that the two maps that should be homotopic are such after post-composition with the inclusion into the total space) Jan 26, 2017 at 23:23

First note that we need a more strict definition of an H-map. An H-map $(f,F):(X,\mu_X)\rightarrow (Y,\mu_Y)$ is a map $f:X\rightarrow Y$ and a homotopy $F:X\times X\rightarrow Y^I$ relative $X\vee X$, satisfying $e_0\circ F=f\circ\mu_x$ and $e_1\circ F=\mu_Y\circ (f\times f)$, where for $a=0,1$, the map $e_a:Y^I\rightarrow Y$ is the evaluation at $a$.
Now the homotopy fibre $F_f$ of $f$ is the (categorical) pullback of $(X\xrightarrow{f} Y\xleftarrow{e_0} PY)$ where $PY=\{l:[0,1]\rightarrow Y\,|\,l(1)=\ast \}$ is the path space over $Y$. That is, $F_f=\{(x,l)\in X\times PY\,|\,f(x)=l(0)\}$. Define $\mu_f:F_f\times F_f\rightarrow F_f$ by
$\mu_f\left((x,l),(y,m)\right)=\left(\mu_X(x,y),F(x,y)+P\mu_Y(l,m)\right)$
It is straightforward to check that this gives a well defined multiplication on $F_f$. Moreover the canonical projection $F_f\rightarrow X$ and the fibre inclusion $\Omega Y\rightarrow F_f$ are H-maps. The multiplication $\mu_f$ depends on the choice of homotopy $F$. If $\mu_X$, $\mu_Y$ have other properties (homotopy associativity, homotopy commutativity, etc) and $f$ preserves these, then often $\mu_f$ too can be shown to inherit these properties too.
• Thanks for your reply. I am not entirely convinced though, for some reasons. The first being the fact that to even say "F is rel $X \vee X$ " we need to rigidify the unit condition (we know it's always doable by choosing a suitable operation homotopic to the given one), but this would not work in the presence of more structure (say homotopies witnessing associativity). Secondly, the ad-hoc nature of the argument does not (imho) look like the right thing to do. Finally, I managed to get the operation myself, but then how do you prove unitality? Jan 27, 2017 at 1:34
• You define an H-space as a unital magma in the based category $ho\mathcal{T}_*$, so the unit should be strict by that definition and $X$, $Y$ should be based by their units by definition. Likewise the H-map $f$ should be based so the relative $X\vee X$ condition on $F$ makes sense. When I say that $F_f$ will inherit any further properties if $f$ preserves them, I really mean that $f$ is given the additional structure of, for example, an $A_n$- or $C_n$-form rather than just the H-map structure $F$. Jan 27, 2017 at 17:28