# Homotopy pushout independent of factorization and symmetric in cofibration category

$$\require{AMScd}$$In Algebraic homotopy, Baues defines the notion of homotopy pushout in a cofibration category in the following way: a commutative diagram

$$\begin{CD} A @>k>> C \\ @AfAA @AAhA\\ B @>g>> D \end{CD}$$

is a homotopy pushout if for one factorization $$B\hookrightarrow W\stackrel{\sim}\to A$$ the induced map $$W\cup_B D\to C$$ is a weak equivalence. He then says "This easily implies that for any factorization $$B\hookrightarrow V\stackrel{\sim}\to A$$ of $$f$$, the map $$V\cup_B D\to C$$ is a weak equivalence. Thus in the definition we could have replaced 'some' by 'any' or used $$g$$ in place of $$f$$." This is page 9 of the book, and no theory has been developed yet, except the construction of cylinders.

Q1. "If one factorization works, all of them do." I managed to reduce this to the following: suppose in addition that there is trivial cofibration $$W\to V$$ that is a 'morphism of factorizations' in the obvious sense. One can then carry out the pushouts for both factorizations and get an induced map from $$W\cup_D B\to V\cup_D B$$. Unless I am missing something this is a pushout of $$W\to V$$, and then a weak equivalence. If this is true, I have a proof, since by taking pushouts and factoring I can connect any two factorizations by a diamond of factorizations $$W\to Q\leftarrow V$$ where both maps are trivial cofibrations. Is there a simpler way to proceed?

Q2. "One can resolve the other variable". It also says "...or used $$g$$ in place of $$f$$." Does the phrasing suggest this is a consequence of the fact any factorization for $$f$$ works if one does? This is not clear to me. I think an argument similar to the one for Q1 should work in proving one can connect everything using a pushout of two factorizations (one of $$f$$ and one of $$g$$), much like one does when balancing Tor or Ext.

Edit. Answer to Q2 is "yes": This follows easily by using Q1 and taking a few pushouts and using C1 and C2.

For completeness, here are the axioms of a cofibration category, which is endowed with a class of cofibrations and of weak equivalences.

C1. Cofibrations and weak equivalences contain the isomorphisms, weak equivalences satisfy 2 out of 3 and cofibrations are closed under compositions.

C2. Pushouts exist along cofibrations, and cofibrations are stable under them. Moreover, weak equivalences are stable under pushouts along cofibrations in both directions.

C3. Maps can be factored into a cofibration followed by a weak equivalence.

C4. Every object admits a trivial cofibration into an object $$R$$ which is fibrant, in the sense every trivial cofibration $$R\to Q$$ splits.

## 2 Answers

I don't remember how Baues does this exactly, but all facts of this sort follow from the Gluing Lemma (see Lemma 1.4.1 in this paper) and "Brown type factorization". By this I mean the following construction. Given a morphism $X \to Y$ and two factorizations $X \to Z_0 \to Y$ and $X \to Z_1 \to Y$ (as cofibrations followed by weak equivalences), we factor the induced morphism $Z_0 \sqcup_X Z_1 \to Y$ as $Z_0 \sqcup_X Z_1 \to Z_2 \to Y$ and obtain a factorization $X \to Z_2 \to Y$ that is weakly equivalent to both original ones. If you do that to a map in a span, you will obtain a zig-zag of weak equivalences connecting spans obtained from any two factorizations and you can apply the Gluing Lemma to that. (Note that axiom C4 is never used in this argument.)

• Yes, the first construction you're mentioning is what I did for Q1 (although perhaps my wording is a bit unclear?). One only needs C1, C2 and C3. Thanks for the reference, by the way. I suppose then that the answer to "Is there a simpler way to proceed?" is "Well, this is the usual way to proceed anyways." – Pedro Tamaroff Jul 17 '18 at 14:22
• Right, I don't think there is an easier way. But I should add that you need to assume that objects $A$, $B$ and $D$ are cofibrant. – Karol Szumiło Jul 18 '18 at 8:29

(The following was intended as a comment to Karol's answer, that after using the "Brown Type Factorization" trick, we can prove the result by applying a result in the book, which does not assume cofibrantness, but due to the word count, I think it might be more appropriate to present it as an answer.)

First, using the Brown Type Factorization trick, we can reduce to showing the following case : suppose the factorization involving $$W,V$$ are weakly equivalent, that is, if we have a diagram :

\begin{align}B\rightarrow V\rightarrow A\\||\quad\;\downarrow i \;\quad||\\B\rightarrow W\rightarrow A\end{align}

(with the middle horizontal arrow being a weak equivalence)

then $$W\cup_BD\to C$$ is a weak equivalence iff $$V\cup_BD\to C$$ is a weak equivalence.

By (C1), it then suffices to show $$i\cup_B 1_D:V\cup_BD\to W\cup_B D$$ is a weak equivalence.

Now the result follows from Lemma 1.2(b) in chapter II in the book applied to the diagram

\begin{align}V\leftarrow B\rightarrow D\\i\downarrow\quad\;||\quad\quad||\\W\leftarrow B\rightarrow D\end{align}