# one question about homotopy pushout

This question arises when I'm reading Jacob Lurie's Higher Topos Theory, p814.

Suppose we are given a diagram $$A_0\leftarrow A\rightarrow A_1$$ in a model category $\mathcal C$. In general, the pushout $A_0\coprod_AA_1$ is poorly behaved in the sense that a map of diagrams $$\begin{array}{rcl} A_0&\leftarrow& A&\rightarrow &A_1\\ \downarrow&&\downarrow&&\downarrow\\ B_0&\leftarrow&B &\rightarrow &B_1 \end{array}$$ need not induce a weak equivalence $A_0\coprod _AA_1\rightarrow B_0\coprod_BB_1$, even if each of the vertical arrows in the diagram is individually a weak equivalence. To correct this difficulty, it is convenient to introduce the left derived functor of 'pushout'. The homotopy pushout of hthe diagram$$A_0\leftarrow A\rightarrow A_1$$ is defined to be the pushout $A_0'\coprod_{A'}A_1'$, where we have chosen a commutative diagram $$\begin{array}{rcl} A_0'&\xleftarrow{j}& A'&\xrightarrow{i} &A_1'\\ \downarrow&&\downarrow&&\downarrow\\ A_0&\leftarrow&A &\rightarrow &A_1 \end{array}$$ where the vertical maps are weak equivalences, and the top row is cofibrant diagram in the sense that $A'$ is cofibrant and the maps $i$ and $j$ are both cofibrations. One can show that such a diagram exists and the pushout $A_0'\coprod_{A'}A_1'$ depends on the choice of diagram only up to weak equivalence.

My question is, given two such diagrams, how to construct such a weak equivalence between the two pushouts?

• What precisely do you mean? Showing that $A'_0\coprod _{A'} A'_1\rightarrow B'_0\coprod_{B'} B'_1$ is a weak equivalence if all vertical maps in the first diagram are weak equivalences? – Paul Frost Aug 4 '18 at 9:08
• @PaulFrost I mean if we the homotopy pushout defined as above is up to a weak equivalence. – user12580 Aug 5 '18 at 1:30

I’m not sure if I understand it quite right, so any comments are welcome. Indeed, Let $$\mathcal B$$ be the category with three objects $$a$$, $$b$$ and $$c$$ and two non identity morphisms. Then $$\mathcal B$$ is a direct category. Then the functor category $$\mathcal C^{\mathcal B}$$ has the model category structure as shown in Hovey’s book Model Categories Theorem 5.1.3. A cofibrant functor is precisely described by Lurie as above. The colimit functor is a left Quillen functor and there is a trivial fibration between a cofibrant diagram and the original diagram. Then any cofibrant resolution of the original diagram is weak equivalent to this cofibrant diagram. Hence any two cofibrant resolution is connected by a Zia-Zag of weak equivalences. However, I can't prove that they can be connected directly by one weak equivalence.