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What on Earth is a homotopy pullback of $$A \rightarrow B \leftarrow C \ \ \ \ \ ???$$ Here $A,B,C$ are elements of a category ${\mathcal V}$ enriched in topological spaces (any convenient category of topological spaces will do). I understand that it is some kind of a weighted limit. This means that I need to take all objects of my category $X$, consider all pullback diagrams $${\mathcal V}(X,A) \rightarrow {\mathcal V}(X,B) \leftarrow {\mathcal V}(X,C)$$ and take their homotopy pullbacks $${\mathcal V}(X,A) \times^{h}_{{\mathcal V}(X,B)} {\mathcal V}(X,C) \; .$$ Now the homotopy pullback is just an object $D$ representing this functor, up to weak homotopy equivalence: $${\mathcal V}(X,D) \ \ \ \stackrel{h}{\sim} \ \ \ {\mathcal V}(X,A) \times^{h}_{{\mathcal V}(X,B)} {\mathcal V}(X,C) \; .$$ This is all grand and dandy but how under the Moon can I get my teeth into this object? Are there any explicit ways of calculating it, similar to the methods, working for the bog down homotopy pullbacks and pushout?

Please, do not try to put any model category structure on ${\mathcal V}$. I am semi-clear how to proceed in the case when one can do it.

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    $\begingroup$ Do you mean pullback ? $\endgroup$
    – Maxime Ramzi
    Commented Mar 10, 2021 at 21:18
  • $\begingroup$ Yes, sorry. I will correct. But the problem is valid for both... $\endgroup$
    – Bugs Bunny
    Commented Mar 10, 2021 at 21:22
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    $\begingroup$ Well very often you can replace your diagram, up to equivalence, by a "good" diagram where the pullback is also a homotopy pullback. This is one of the raisons d'être of model structures, although you need much less than that. For instance if you can find a diagram $A' \to B' \leftarrow C'$ which is related to your original diagram by a zigzag of maps of diagrams, each of which inducing equivalences on mapping spaces, and such that $V(X,C')\to V(X,B')$ is a fibration for all $X$, then you can just take the pullback $A'\times_{B'} C'$. $\endgroup$
    – Maxime Ramzi
    Commented Mar 10, 2021 at 21:31
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    $\begingroup$ But ultimately it will depend on your specific situation $\endgroup$
    – Maxime Ramzi
    Commented Mar 10, 2021 at 21:32
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    $\begingroup$ If your category is cotensored over spaces and has pullbacks, you can often construct a homotopy pullback as the fiber product $(A \times C) \times_{B \times B} B^{[0,1]}$ (i.e. apply the "classical" homotopy pullback formula from spaces). Anything isomorphic to that in the homotopy category of $\mathcal{V}$ will also be a homotopy pullback. $\endgroup$
    – Tyler Lawson
    Commented Mar 10, 2021 at 21:50

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Are there any explicit ways of calculating it, similar to the methods, working for the bog down homotopy pullbacks and pushout?

Yes, in fact the same formula continues to work in this case.

Consider the (ordinary) pullback $$A⨯_B B^{[0,1]} ⨯_B C.$$

Here $B^{[0,1]}$ denotes the powering of $B$ over $[0,1]$, defined by the following universal property: ${\cal V}(X,B^{[0,1]})$ is naturally isomorphic to ${\cal V}(X⨯[0,1],B)$.

Now, if we compute ${\cal V}(X,A⨯_B B^{[0,1]} ⨯_B C)$, we get $${\cal V}(X,A)⨯_{{\cal V}(X,B)}{\cal V}(X⨯[0,1],B)⨯_{{\cal V}(X,B)}{\cal V}(X,C),$$ where the middle term ${\cal V}(X⨯[0,1],B)$ is isomorphic to $({\cal V}(X,B))^{[0,1]}$, yielding $${\cal V}(X,A)⨯_{{\cal V}(X,B)}({\cal V}(X,B))^{[0,1]}⨯_{{\cal V}(X,B)}{\cal V}(X,C),$$ which is precisely the usual formula for the homotopy pullback of topological spaces, namely $${\cal V}(X,A)⨯^h_{{\cal V}(X,B)}{\cal V}(X,C).$$

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