Let $$ \begin{array}{ccc} A & \to & B \cr\downarrow&&\downarrow \cr A'& \to &B' \end{array} $$ be a pullback square in the category of all topological spaces (not just in a convenient one) and the arrow $B\rightarrow B'$ a Dold fibration.

Is the square always a homotopy pullback square

  1. in the Quillen model structure?
  2. in the Strom model structure?

In less abstract terms, is the canonical map from the pullback of the Hurewicz fibration associated to $B\rightarrow B'$ with $A'\rightarrow B'$ to $A$ a

  1. weak homotopy equivalence?
  2. homotopy equivalence?

Note that a Dold fibration may not be a Serre fibration.

  • $\begingroup$ It seems potentially likely in the Strøm model structure. Note you should say a Serre fibration might not be a Dold fibration (not so much the other way around), but a Hurewicz fibration is a Dold fibration (see mathoverflow.net/questions/20442/…) $\endgroup$
    – David Roberts
    Commented Aug 3, 2015 at 15:18

1 Answer 1


In this paper of Charles Rezk he proves, among other things, that in any right proper model category, all base changes of a map $f:B \rightarrow B'$ give homotopy Cartesian squares if and only if the map is sharp.

Definition. A map $f:X \rightarrow S$ is sharp if, after any base change to $X'\rightarrow S'$ and weak equivalence $S''\rightarrow S'$ , the map $X'\times_{S'}S''\rightarrow X'$ is also a weak equivalence.

Dold fibrations are stable under pullback so it suffices to check the condition about weak equivalences for a Dold fibration. If your weak equivalences are homotopy equivalences then this is true. The Strøm model structure is proper (since everything is fibrant and cofibrant), and the weak equivalences are homotopy equivalences so this answers one question.

I don't know what happens in the Serre or Hurewicz model structures, bot both are right proper so you're reduced to checking the stability of pulling back weak equivalences against a Dold fibration, which is presumably easier.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.