Direct proof that the model category of cdgas is left proper Let $k$ be a field of characteristic $0$. The projective model structure on the category $cdga$ of commutative differential graded $k$-algebras is proper. Since this model structure is transferred from the projective model structure on chain complexes, it follows formally that the projective model structure on $cdga$ is right proper. However the only reference for it being left proper I know is Toen &  Vezzosi's HAG II, and this proof is roundabout, in that they show that $cdga$ forms a HAG context, which hence implies it must be left proper. Does someone know a more direct proof, or a reference with one? Thanks!
 A: If you want a reference other than Toen-Vezzosi, this statement is proven as Theorem 4.17 in my paper Commutative Monoids in General Model Categories, plus example 5.1 to see that the commutative monoid axiom holds for chain complexes in characteristic zero, plus the table on page 3 of Batanin-Berger Homotopy theory for algebras over polynomial monads, which checks that chain complexes in characteristic zero are h-monoidal and compactly generated. By the way, this proof specializes to Tyler's proof (i.e. using the same diagram plus the same filtration), but is phrased to hold in much more general settings. My proof is just a riff on the similar proof which is one of the main results in the Batanin-Berger paper.
A: The model structure on this category has a set of generating cofibrations: if we write $F$ for the free cdga functor, $k$ for the complex with value $k$ concentrated in degree zero, and $I$ for the mapping cone of the identity $k \to k$, then the maps $F(k[n]) \to F(I[n])$ form a set of generating cofibrations. Weak equivalences are preserved by filtered colimits in this category, so to prove that the model category is left proper, it suffices then to show that in any double pushout diagram
$$
\require{AMScd}
\begin{CD}
F(k[n]) @>>> A @>{\sim}>> B \\
@VVV @VVV @VVV\\
F(I[n]) @>>> A' @>>> B',
\end{CD}
$$
we have that the map $A' \to B'$ is a quasi-isomorphism. (You show that the set of maps $f$ such that pushing out along $f$ preserves quasi-isos is closed under retracts, cobase change, and transfinite composition, and the above shows that it contains the generating cofibrations; that means that it contains all the cofibrations.)
Equivalently, given $f: A \to B$ a quasi-isomorphism of cdgas and a cycle $\alpha \in A_{k-1}$, we need to show that the map $A [x] \to B[x]$ is a quasi-isomorphism, where $x$ is a new free generator in degree $k$ with boundary $\alpha$ (in particular, $x$ squares to zero if it is in odd degree).
We have compatible, exhaustive filtrations of these algebras by the subcomplexes $F_p A[x] = A \cdot \{1, x, \ldots, x^p\}$ and similarly $F_p B[x]$. The map of subquotients $F_p A[x] / F_{p-1} A[x] \to F_p B[x] / F_{p-1} B[x]$ is isomorphic to either the map $A[kp] \to B[kp]$ or $0 \to 0$ (depending on whether $x^p$ is zero or not). Since $A \to B$ is a quasi-isomorphism, the map of subquotients is also quasi-isomorphism. By induction we find that $F_p A[x] \to F_p B[x]$ is a quasi-isomorphism and hence so is $A[x] \to B[x]$.
(Oddly, the proof doesn't depend on the model structure existing.)
