I'm reading a book on homotopical algebra, and trying my best to understand the various proofs etc but I'm new to model categories and weak factorisation systems, so there are a lot of details I'm struggling with, which I assume are routine/easy for the more experienced reader. First let me state the lemma :
Ken Brown's Lemma: Let $C$ be a model category, together with a functor $F:C\to D$. Assume that $D$ is endowed with a class of weak equivalences, by which we mean a class of morphisms which contains all isomorphisms and which has the 2-out-of-3 property. If $F$ sends trivial cofibrations between cofibrant objects to weak equivalences, then it sends weak equivalences between cofibrant objects to weak equivalences.
The proof goes as follows :
Let $f:X\to Y$ ve a weak equivalence between cofibrant objects. We may form the push out square $\require{AMScd}$
\begin{CD}
\emptyset @>>>Y \\
@VVV @VVjV \\
X@>>i> X\coprod Y
\end{CD}
and see that, since $X$ and $Y$ are cofibrant, the canonical maps from $X$ and $Y$ to the coproduct $X\coprod Y$ are cofibrations.
This is where the trouble begins for me (very early on as you can see). I've tried showing that $i,j$ are indeed cofibrations, but I have not succeeded. The two ways I thought might work was either showing that $i$ had the left lifting property with respect to the trivial fibrations, since $Cof=l(W\cap Fib)$ this would do the trick. My other guess was showing that $i$ was the retract of some cofibration but I couldn't do it either. So this is the first question, why is it that the canonical maps $X,Y\to X\coprod Y$ are cofibrations, given that $X,Y$ are cofibrant ?
We may factor the map $(f,1_Y):X\coprod Y\to Y$ into a cofibration $k:X\coprod Y\to T$ followed by a weak equivalence (trivial fibration) $p:T\to Y$. We thus have two commutative triangles : (drawn as squares here for convenience)
$\require{AMScd}$
\begin{CD}
X @>>ki> T @.@. Y @>>kj> T\\
@VVfV @VVpV @. @V1_YVV @VpVV \\
Y @= Y @.@. Y @= Y
\end{CD}
Since the map $F(p)$ has a section which is the image of a trivial cofibration between cofibrant objects, it is a weak equivalence.
This is second place where I'm having issues. What's the section of $F(p)$ ? It is not even clear what is meant by section here, as the word is used for various different things up to that point in the book, none of which seems particularly fitting for the context. I think it must be something like $s$ is a section of $r$ if $rs=1$. Because then $p\circ kj=1_Y$, hence the section is $F(kj)$. So, if this is true, the proof says $kj$ is a trivial cofibration between cofibrant objects. $Y$ is cofibrant by assumption, but I don't know why $T$ should be cofibrant. So 2nd question is, am I correct about the section ? If so, why is $kj$ a trivial cofibration between cofibrant objects ?
On the other hand the map $ki$ is a trivial cofibration between cofibrant objects and therefore, its image $F(ki)$ is a weak equivalence. The 2-out-of-3 property in $D$ implies that $F(f)$ is a weak equivalence.
Third and last question is similar to the previous one, why is $ki$ a trivial cofibration ? Again, I have a few ideas on how one could prove this as in the first question but I can't actually solve my issues.
Best Answer
It's a basic result that the pushout of a cofibration is a cofibration. This would show $i,j$ to be cofibrations.
$\require{AMScd}$ $$\begin{CD}A@>f>>B\\@VgVV@VVhV\\C@>>i>D\end{CD}\quad\quad\quad\begin{CD}B@>u>>E\\@VhVV@VVvV\\D@>>w>F\end{CD}$$
Let the leftmost square be a pushout where $g$ is a fibration; let the second square be any commutative square where $v$ is a trivial cofibration. We want to demonstrate a lift $\phi:D\to E$. Well, we have the square: $$\begin{CD}A@>uf>>E\\@VgVV@VVvV\\C@>>wi>F\end{CD}$$So that there is a lift $\psi:C\to E$ making both triangles commute - $g$ is a cofibration. Now $\psi,u$ make a cocone under the pushout diagram - with nadir $E$ - (because $\psi g=uf$) so that there is a (unique) $\phi:D\to E$ with $\psi=\phi i$ and $u=\phi h$. It remains only to check that $v\phi=w$: well, both $v\phi$ and $w:D\to F$ have the same components $C\to F,B\to F$ so by the uniqueness property of a pushout they must be the same.
Thus, such a lift $\phi:D\to E$ exists (and is unique if $\psi$ is) so that $h$ - the pushout of $g$ - is a cofibration.
The "section" would be $F(kj):F(Y)\to F(T)$ (because $F(p)\circ F(kj)=\mathrm{Id}$). $kj$ is a trivial cofibration between cofibrant objects, so $F(kj)$ is a weak equivalence - by hypothesis - so that it follows by the $2$-of-$3$ that $F(p)$ is also a weak equivalence.
$T$ is cofibrant because the unique map $\emptyset\to T$ equals the composite $\emptyset\to X\overset{i}{\longrightarrow}X\sqcup Y\overset{k}{\longrightarrow}T$ which is a composite of cofibrations hence a cofibration. $kj$ is a trivial cofibration because it is a cofibration and the diagram $p(kj)=\mathrm{Id}$ - with $p$ a weak equivalence - implies $kj$ to be a weak equivalence.
$ki$ is a weak equivalence for the same reason - the $2$-of-$3$ applied to $p=f(ki)$ - and is a composite of cofibrations, hence a cofibration.