A Mayer-Vietoris argument I don't get -- From a paper by Bousfield-Gugenheim I working through the article On PL De Rham theory and rational homotopy type by Bousfield-Gugenheim. I am having troubles completing the proof of lemma 5.3, where they claim that the fact that a certain map is a weak equivalence follows from a Mayer-Vietoris argument.

Some background:
Let $\mathcal{A}$ be the category of dg commutative algebras in non-negative grading, $\mathcal{S}$ the category of simplicial sets, $\nabla(n,*)$ the Sullivan algebra (it is a simplicial dg commutative algebra; we won't need the precise definition here). We have a contravariant functor
$$A:\mathcal{S}\longrightarrow\mathcal{A}$$
given by $A(K)_n = \mathcal{S}(K,\nabla(n,*))$ with differential and multiplication induced by those of $\nabla(n,*)$.
There is a closed model structure on $\mathcal{A}$ where the fibrations are the surjections, the weak equivalences are the quasi-isomorphisms, and the cofibrations are defined by lifting property.

My problem:
Let $p:X\to Y$ be a fibration in $\mathcal{A}$, and $u:K\to L$ an inclusion of finite simplicial sets which is also a weak equivalence. The claim is that the map
$$\phi:=(A(u)\otimes1,1\otimes p):A(L)\otimes X\longrightarrow (A(K)\otimes X)\times_{A(K)\otimes Y}(A(L)\otimes Y)=: Q$$
is a trivial fibration.
The fact that it is surjective (i.e. a fibration) follows trivially from the fact that $p$ is surjective, and that by lemma 2.7 in the paper $A(u)$ also is. It is left to show that $\phi$ is a quasi-isomorphism. They say that it follows easily from a Mayer-Vietoris argument, so I presume that I have to cook up a short exact sequence in $\mathcal{A}$, one of the maps being $\phi$ and one object possibly having trivial homology, but I don't really see how. is there a standard way?
Of course, the obvious SES would be
$$0\xrightarrow{}\ker(\phi)\hookrightarrow{}A(L)\otimes X\xrightarrow{\phi}Q\xrightarrow{}0,$$
but in this case, how do I characterize the kernel of $\phi$?
I believe
$$\ker(\phi) = (\ker(A(u))\otimes X)\sqcup(A(L)\otimes\ker(p)),$$
but I don't see why this should be acyclic...

I realized that I confused Mayer-Vietoris with the LES induced by a SES. A possible algebraic statement for Mayer-Vietoris I was able to derive (but I didn't check all the details) is the following:
Consider a push-out square
$$\require{AMScd}
\begin{CD}
U @>i>> W\\
@VjVV @VVkV\\
V @>>h> M
\end{CD}$$
in the category of cochain complexes, $M= V\sqcup_{U}W$, and assume that $i$ is injective. Then we have a long exact sequence
$$\ldots\longrightarrow H^n(U)\xrightarrow{i_*\oplus j_*} H^n(V)\oplus H^n(W)\xrightarrow{h_*-k_*}H^n(M)\stackrel{\delta}{\longrightarrow}H^{n+1}(U)\ldots$$
The boundary map is defined as follows: let $[\alpha]\in H^n(M)$. As the map
$$h+k:V\oplus W\longrightarrow M$$
is surjective, we have $\alpha^V\in V$ and $\alpha^W\in W$ mapping to $\alpha$. Therefore,
$$0 = d\alpha = h(d\alpha^V) - k(d\alpha^W).$$
It follows that there exists $\overline{\beta}\in U$ such that $i(\overline{\beta}) = d\beta^V$ and $j(\overline{\beta}) = d\beta^W$. As $i(d\overline{\beta}) = di(\overline{\beta}) = 0$, injectivity of $i$ implies that $\overline{\beta}$ is closed, so that we can define $\delta[\beta] = [\overline{\beta}]$.
Now the two questions I have left are:


*

*How to "dualize" this argument and get a Mayer-Vietoris sequence associated to a pull-back square?

*How to apply this to the specific situation above? It is not a pull-back square, but a map induced by universal property of one...
 A: I just had the same problem and tried to solve it on my own, I think my solution works but I don't think it can be called a "Mayer-Vietoris" argument.
The proof relies on the following lemma, that you can find proven in Jardine's famous online Lectures on homotopical algebra https://www.uwo.ca/math/faculty/jardine/courses/homth/homth-lecture007a.pdf
as lemma 17.3.
Basically, this lemma states  that, in any model category,the pullback of a weak equivalence along a fibration is again a weak equivalence, if all the object are fibrant.
Then, if one takes the pull-back along $Au\otimes 1$ and $1\otimes p$, that is, the diagram
\require{AMScd}
\begin{CD}
AK\otimes X \times_{AK\otimes Y} AL\otimes Y @> >> AL\otimes Y\\
@V VV @VV Au\otimes 1V\\
AK\otimes X @>>1\otimes p> AK\otimes Y
\end{CD}
the vertical map on the left is still going to be a weak equivalence. Then the map $(AU\otimes 1,1\otimes p)$ is a weak equivalence because of the two out of three property, and because the vertical map before and the map $Au\otimes 1$ are (notice that this time I mean the map $Au\otimes 1:AL\otimes X\to AK\otimes X$)   .
