Fiber of fibration of simplicial sets

Question

$\require{AMScd}$ If $p:E\to B$ is a fibration of simplicial sets, is the fiber in the model category sense, i.e. the homotopy limit of $$\begin{CD}{} @. E \\@. @VVV \\*@>>> B \end{CD}$$

the same as the set-theoretic fiber $p^{-1}(*)$ (where $*$ is some zero-simplex of $B$)?

I feel like this is supposed to be obvious, but I'm confused by homotopy limits: you're supposed to fibrantly replace everything, and then take the categorical limit, right? If $E$ and $B$ aren't fibrant, I don't have a good down-to-earth model for the fibrant replacement, so I'm kind of stuck...

Answer 1

Yes, the homotopy fibre of a Kan fibration (of simplicial sets) coincides with the ordinary fibre. This is not supposed to be obvious: it is a consequence of right-properness, which is in turn a consequence of the existence of a very good fibrant replacement functor.

In general, given morphisms $f : X \to Z$ and $g : Y \to Z$ in some model category (not necessarily right-proper), the homotopy pullback can be computed as follows:

1. Choose a fibrant replacement $Z \to \hat{Z}$.
2. Factorise $X \to Z \to \hat{Z}$ as a weak equivalence followed by a fibration $\hat{X} \to \hat{Z}$.
3. Factorise $Y \to Z \to \hat{Z}$ as a weak equivalence followed by a morphism $\hat{Y} \to \hat{Z}$ where $\hat{Y}$ is fibrant.
4. Then the homotopy pullback is $\hat{X} \times_{\hat{Z}} \hat{Y}$.

In a right-proper model category, you can skip (1) and (3). However, (2) is crucial.

You may also find the description of homotopy pullbacks in $\mathbf{Top}$ given here useful.