Defining homotopy fiber "naturally"

Question

For a map $f\colon A\to B$, many authors define the mapping fibration $E_f\to B$, where $E_f$ consists of pairs $(a,\gamma)\in A\times B^I$ with $\gamma(0)=f(a)$, and the fibration map is $(a,\gamma)\mapsto \gamma(1)$. The fiber $F_f$ is the homotopy fiber, and consists of pairs $(a,\gamma)$ with $\gamma$ a path from $f(a)$ to a basepoint $b\in B$.

At the same time, there are many god-given fibrations lying around that are confusing similar. For example, we have $B^I\to B$, with $\gamma\mapsto \gamma(0)$. As sets, $E_f$ is just the pullback of $f$ along this fibration. However, this doesn't actually produce the right fibration map $E_f\to B$ is any obvious way.

So my (soft) question is the following:

Is there a way to define the homotopy fiber of a map $f\colon A\to B$ which is "natural" in the colloquial sense?

I should also clarify that I don't know any model theory and am asking this as a newcomer to homotopy theory.

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Partial answer:

The most satisfying solution I have found is to start with the path fibration $PB$ consisting of paths starting at a basepoint $b_0$. The map $PB\to B$ with $\gamma\mapsto g(1)$ is a fibration, and $F_f$ is the pullback of $f\colon A\to B$ along this. This is almost satisfying, but still requires picking this evaluation map.

Answer 1

Tldr; jump to the $(\ast)$.

The category $Top$ of topological spaces is complete and cocomplete. That is, all limits and colimits exist in the category. On the other hand, the homotopy category $hTop$, formed by quotienting out the (weak) equivalences, is neither complete nor cocomplete. In fact while it has products and coproducts, it has very few other limits or colimits. See here for some examples of pullback/pushout squares which do not exit in $hTop$.

The point is that the strict fibre of an arbitrary map $f:X\rightarrow Y$ over a point $y\in Y$ is the equaliser of the two maps $f,c_y:X\rightrightarrows Y$, where $c_y$ is the constant map at $y$. Now this is not a good homotopical notion, in general, due to the lack of limits in $hTop$. It is a theorem that if a category has equalizers and finite products then it has all finite limits, and we have already noted above that $hTop$ does not even have all pullbacks. Since it has products $hTop$ cannot have all equalisers.

To see exactly what goes wrong consider the fact that the strict fibres of $f:X\rightarrow Y$ over the different points of $Y$ will not in general have the same homotopy type (even keeping within the same path component). For instance consider the map $S^1\rightarrow \mathbb{R}$ induced by projecting onto the first coordinate; the fibres are either empty, have one point, or have two points.

A sufficient condition that all the strict fibres over all points in a given path component have the same homotopy type is that the map $f$ is a fibration. And this is what the homotopy fibre does: it replaces the map $f$ by a fibration $p_f:E(f)\rightarrow Y$ and then takes its strict fibre over a given point. For everything to make sense we require some comparison map $j_f:X\rightarrow E(f)$ which is a homotopy, or at least weak, equivalence, and that everything should be natural - in the mathematical sense - at least up to homotopy.

Note that this 'up-to-homotopy' comparison map $j_f$ is the best we can do, for if $j_f$ were a homeomorphism, then the map $f$ would already be a fibration, and its homotopy fibres would just be its strict fibres. Thus if it were always possible to replace a map by a fibration up to homeomorphism, then there would be no need for homotopy fibres to begin with.

$(\ast)$ Now, the point is that the homotopy fibre of a map $f:X\rightarrow Y$ does not really live in $Top$. Since at best we can ask for the map $j_f$ to be a homotopy equivalence, the homotopy fibre of $f$ really lives in the homotopy category $hTop$ where it makes sense to consider the objects $X$ and $E(f)$ as the same. And now in the homotopy category everything is perfectly canonical, since $X$ and $E(f)$ are the same object as seen through categorical eyes. For example in the path space fibration we no longer have to pick an evaluation, since all these maps are homotopy equivalent, and thus represent the same morphism in $hTop$.

Thus we realise that the object $E(f)$ is nothing but an auxiliary construct. It is a point set lift of the problem in $hTop$ to a problem in $Top$. It allows us to create a more understandable model for the problem by using spaces with points, and maps defined on elements, rather than having everything 'up-to-homotopy'. However there is no canonical way to achieve this. For, just as the strict fibres of an arbitrary map $f:X\rightarrow Y$ may be badly behaved, the fibres of the projection $Top\rightarrow hTop$ are not uniform. Asking for a canonical section of this map is too much.

Thus we content ourselves with mathematically natural (again, up to homotopy) constructions point-set constructions, happy with the knowledge that once we project the problem back to $hTop$ we truly do have something safe and canonical.

Answer 2

A general philosophy is that you want "equivalent" (in a suitable homotopical sense), not "equal".

The strict notion of the fiber over the basepoint $b$ of a map $f : A \to B$ is that it is a space whose points correspond to having a point $a \in A$ together with an assertion that $f(a) = b$.

In homotopy theory, that auxiliary data should consist of an equivalence $f(a) \simeq b$ rather than asserting an equation, or even merely asserting the proposition that an equivalence exists. So, in a formal sense, the points of the homotopy fiber over $b$ should consist of pairs $(a,p)$ where $a$ is a point of $A$ and $p$ is an equivalence between $f(a) \simeq b$.

The typical way of expressing an equivalence is as a path: given $b, b' \in B$, an equivalence from $b$ to $b'$ is given by a path $p : I \to B$ with $p(0) = b$ and $p(1) = b'$.

This construction topologizes in a natural way; we should get a model for the homotopy fiber over $b$ by taking the subspace of $E_f \subseteq A \times B^I$ comprised of pairs $(a,p)$ such that $f(a) = p(0)$ and $b = p(1)$.

I am, of course, glossing over the work in elaborating if and how this sort of thing works coherently to describe the sorts of things we intend to describe; I hope, however, it's enough to get an idea about the motivation.

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As a systematic construction, you should consider the fibration $B^I \to B \times B$ (whose components are the "start point" and "end point" maps) as a topological object representing all the different equivalences possible between two points of $B$.

In particular, given any point $(b, b') \in B \times B$, the (strict) fiber over that point represents the space of all equivalences $b \simeq b'$. Topologically, of course, this is the space of all paths (with fixed endpoints) from $b$ to $b'$.

So, given variables $a \in A$ and $b \in B$, the way to express $f(a) \simeq b$ is to take the pullback bundle

$$ \require{AMScd} \begin{CD} M @>>> B^I \\ @VVV @VVV \\ A \times B @> f \times 1_B>> B \times B \end{CD} $$

Then the (strict) fiber of $M \to A \times B$ over any specific point $(a_0,b_0)$ is precisely the space of all equivalences $f(a_0) \simeq b_0$. And if you fix a specific $b$, this construction becomes precisely the one discussed above

$$ \require{AMScd} \begin{CD} E_f @>>> B^I \\ @VVV @VVV \\ A @> f \times \{b\}>> B \times B \end{CD} $$

Your example comes from fixing one of the endpoints as an intermediate step: we have three pullback squares

$$ \require{AMScd} \begin{CD} F_f @>>> PB @>>> B^I \\ @VVV @VVV @VVV \\ A @>f>> B @> \{ b \} \times 1_B >> B \end{CD} $$

The strict fiber of $PB \to B$ over $b' \in B$ is the space of equivalences $b \simeq b'$, and then the strict fiber of $F_f \to A$ over $a \in A$ is the space of equivalences $b \simeq f(a)$.