What does it mean to speak of a homotopy fibration sequence?

Question

I'm reading a paper in which the following is done. We have a certain particular map of spaces $f:X\to Y$ and then it is said something along the lines of "let $Z_f$ denote the space whose defining attribute is that it is the space that fits into the following fibration sequence up to homotopy":

Then it is said "To speak of the sequence above as a homotopy (co)fibration sequence, one needs to specify a preferred null-homotopy for the composite map".

(the (co) is mine, since in fact the writer is working with spectra, and in my interpretation they are being replaced by spaces, where no longer cofibration=fibration)

I don't really understand what's going on. I'm aware of the general fact that the nullhomotopies of $gf$ in the following diagram are in one to one correspondence with the extensions $G$ of $g$ to the homotopy cofiber:

Returning to the case above, why is it needed to specify a preferred null-homotopy for the composite map? Can't we just specify a map $Y\to Z_f$ and prove that the composition is nullhomotopic?

Answer 1

Good definition: $X\to Y\to Z$ is a homotopy fibration sequence if the composed map is a constant $z$ and the resulting map from $X$ to the homotopy fiber of $Y\to Z$ is a weak homotopy equivalence.

More general good definition: $X\to Y\to Z$ is a homotopy fibration sequence if a homotopy is given from the composed map to a constant $z$ and the resulting map from $X$ to the homotopy fiber of $Y\to Z$ over $z$ is a weak homotopy equivalence.

Bad definition: $X\to Y\to Z$ is a homotopy fibration sequence if there is a homotopy from the composed map to a constant $z$ such that the resulting map from $X$ to the homotopy fiber of $Y\to Z$ over $z$ is a weak homotopy equivalence.

What's bad, or at least potentially confusing, about this is that you can have two different homotopies from the composed map to a constant such that one does the job and the other does not.

Answer 2

I have no idea whose paper the question refers to, but I can say what standard naive practice dictates. For any map $f\colon X\to Y$ of based spaces or spectra, there is a canonical cofiber $i\colon Y\to Cf$, where $Cf = Y\cup_f CX$. Here $CX$ is the (reduced) cone $X\wedge I$, where $I$ has basepoint $1$. Thus $Cf$ is the pushout of $f$ and the inclusion $i_0\colon X\to CX$. That is the space (or spectrum, being sloppy about the distinction between fiber and cofiber sequences there) that a working mathematician has in mind for your $Z_f$. However, I emphasize that this is not just a canonical space but a canonical map.

For a map $g\colon Y \to Z$, the resulting sequence of maps $X \to Y \to Z$ is said to be a cofiber sequence if there is an equivalence to the sequence $X \to Y \to Cf$. If we require a map of sequences giving the equivalence to be the identity on $X$ and $Y$, then it is given by a map $h\colon Z\to Cf$ such that $h\circ g = i$. Then $h$ is given by a null homotopy of $g\circ f$, and that may be the "preferred" null homotopy your author has in mind.

This is analogous to algebra, where a cokernel of a homomorphism of groups is not a group but a homomorphism of groups. The analogy becomes precise if one works with chain complexes instead of groups.

Answer 3

I cannot answer more precisely without knowing which paper you are referring to. However the point seems to be that you want to characterize the space $Z_f$. In fact there are plenty of spaces $W$ with a map $h:Y\to W$ such that $hf$ is nullhomotopic (for instance any space $W$ with a nullhomotopic map $h$ will do). Moreover the cofiber is not the universal space with an arrow $h:Y\to W$ such that $hf$ is nullhomotopic: I am not even certain that such a thing exists. As you noticed in the OP the universal property of the cofiber is to be the universal space $W$ with an arrow $h:Y\to W$ and a nullhomotopy $H:hf\sim *$. Which nullhomotopy it is is important for the characterization.

In general in homotopy theory remembering the homotopy between things (and not just that they exists) is an important detail for the statement and the proof of many theorems.

Answer 4

Let me address your last question, and say what goes wrong if we do not prefer a particular nullhomotopy, without saying why that is an adequate solution.

One thing we want out of a homotopy fiber sequence is a long exact sequence of homotopy groups. If we have a sequence $X\to Y \to Z$, then we can apply the $\pi_n$ functor and get maps of groups, but the boundary map $\pi_nZ\to \pi_{n-1}X$ is the difficult part. The choice of that map depends on the choice of the nullhomotopy. For example, this sequence $K(\mathbb Z,n-1)\to *\to K(\mathbb Z,n)$ is a fragment of a homotopy fiber sequence with boundary map an isomorphism $\mathbb Z=\pi_{n}K(\mathbb Z,n)\to\pi_{n-1}K(\mathbb Z,n-1)=\mathbb Z$. But which isomorphism is it? We must do something to break the symmetry between the two isomorphisms. The nullhomotopy is the extra structure that breaks the symmetry and allows us to build boundary maps.