I have a concrete problem with the homotopy fiber and I am getting lost with the literature. I state my question and, to avoid confusions, I state downwards the definitions I am using.

Let $C$ be a pointed model category. To make things simpler, assume every object is fibrant. Recall that the definition of the homotopy fiber of a morphism $f\colon Y\to X$ is the homotopy pullback of the diagram $$\begin{matrix}& & Y \\ &&\downarrow^f \\ *&\to & X\end{matrix} $$ That is to say, we replace $*\to X$ and $f$ by fibrations (in the category of diagrams) and take the classic pullback.

Now, if I have a map $Z\to X$ that fits in a commutative square $$ \begin{matrix} Z&\to &Y\\ \downarrow &&\downarrow^f\\ *&\to & X. \end{matrix}\quad (1) $$ then it factors through the fiber. More concretely, there is a map $Z\to \mathrm{fib}$ such that the diagram $$ \begin{matrix} Z&\to&\mathrm{fib}&\to &Y\\ &&\downarrow &&\downarrow^f\\ &&*&\to & X. \end{matrix} $$ commutes in $C$ and the composed maps $Z\to *$ and $Z\to X$ are those of (1).

My question: is there an analogue map to the homotopy fiber?, i. e., is there a commutative square in $C$ $$ \begin{matrix} Z&\to&\mathrm{hofib}&\to &Y'\\ &&\downarrow &&\downarrow^{Rf}\\ &&*'&\to & X'. \end{matrix}\quad (2) $$ where $*'\to X'$ and $Rf\colon Y'\to X'$ are the fibrant replacements?

Let me recall the general definition of the homotopy limit. Denote $D$ the above diagram. Then $C^D$ has a model structure. Note $\mathrm{lim}\colon C^D\to C$ the functor taking pullback. The homotopy pullback is the right derived functor of $\mathrm{lim}$ in the sense of Quillen in "Homotopical algebra". Therefore, in order to have such a map I should have a functor $F$ such that for every diagram $d\colon D\to C$ I have an object $F(d)$ with maps as in (1) and that in my concrete diagram gives $Z$. Then I would have a diagram like (2). Right?

I have read in n-category lab that this definition is called global and there is another definition of the homotopy fiber called local and that, under certein hypothesis, they seem coincide. Do this terminology refer to this problem? I am unable to understand the notations in the literature.

  • 2
    $\begingroup$ I don't think assuming every object is cofibrant is so helpful here. Do you mean every object is fibrant? If not, then your first definition is incorrect – you should replace $X$ with a fibrant object. $\endgroup$
    – Zhen Lin
    Oct 20, 2015 at 16:29
  • $\begingroup$ @ZhenLin yes, you are right. Thanks for pointing out the mistake. I edit it now $\endgroup$
    – Tintin
    Oct 20, 2015 at 16:41

1 Answer 1


I work in a general pointed model category. The homotopy fibre of a morphism $f : Y \to X$ can be defined as follows: first, choose a fibrant replacement $w_X : X \to \hat{X}$ and then factor $w_X \circ f : Y \to \hat{X}$ as a weak equivalence $w_Y : Y \to \hat{Y}$ followed by a fibration $\hat{f} : \hat{Y} \to \hat{X}$; then the homotopy fibre (with respect to these choices) is the ordinary fibre of $\hat{f} : \hat{Y} \to \hat{X}$.

Thus, we have the following commutative diagram, $$\require{AMScd} \begin{CD} \operatorname{fib} f @>>> Y @>{w_Y}>> \hat{Y} @<<< \operatorname{hfib} f\\ @VVV \mathrm{pb} @VV{f}V @V{\hat{f}}VV \mathrm{pb} @VVV \\ 0 @>>> X @>>{w_X}> \hat{X} @<<< 0 \end{CD}$$ and in particular, we get a comparison $\operatorname{fib} f \to \operatorname{hfib} f$ making the following diagram commute: $$\begin{CD} \operatorname{fib} f @>>> \operatorname{hfib} f \\ @VVV @VVV \\ Y @>>{w_Y}> \hat{Y} \end{CD}$$

If you can choose $(\hat{X}, w_X, \hat{Y}, w_Y, p)$ functorially, then you get a homotopy fibre functor, and it can be shown to be a right homotopical approximation of the fibre functor. This is the "global" universal property. The "local" universal property is more or less built into the definition I give above. One way of getting a feel for it is to think about what happens at the level of the homotopy category: in this case, it says that, for any morphism $g : Z \to Y$ such that $f \circ g = 0$ in the homotopy category, there is a (not necessarily unique) morphism $Z \to \operatorname{hfib} f$ making the evident diagram commute (in the homotopy category). But, of course, this is not a complete characterisation of $\operatorname{hfib} f$.

  • $\begingroup$ Thank you very much for your clear answer, Zhen. It is very well explained and written. $\endgroup$
    – Tintin
    Oct 25, 2015 at 11:35

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