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!

1$\begingroup$ I think there is a proof in Baues's algebraic homotopy book. If so, her probably assumes connectivity, but it is unnecessary, as Tyler shows below. $\endgroup$ – Fernando Muro May 1 '15 at 6:27
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 quasiisomorphism. (You show that the set of maps $f$ such that pushing out along $f$ preserves quasiisos 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 quasiisomorphism of cdgas and a cycle $\alpha \in A_{k1}$, we need to show that the map $A [x] \to B[x]$ is a quasiisomorphism, 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 $F_p A[x] / F_{p1} A[x] \to F_p B[x] / F_{p1} B[x]$ 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 quasiisomorphism, the map of subquotients is also quasiisomorphism. By induction we find that $F_p A[x] \to F_p B[x]$ is a quasiisomorphism and hence so is $A[x] \to B[x]$.
(Oddly, the proof doesn't depend on the model structure existing.)

1$\begingroup$ The notion of left properness (and also right properness) only depends on the class of weak equivalences and not on the model structure (or existence thereof), so it's not surprising that the proof makes no use of a model structure. $\endgroup$ – Dmitri Pavlov Jun 16 '15 at 13:25

$\begingroup$ This is for unbounded dgcas, right? Does this also go through for the category of cochain dgcalgebras in nonnegative degrees (the one used in Sullivan style rational homotopy theory)? This has the same generating cofibrations in positive degree, but then in addition the map from zero to the free graded algebra on a single generator in degree 0 and, curiously, also the map going the other way around. $\endgroup$ – Urs Schreiber Feb 20 '17 at 18:29

1$\begingroup$ @Urs I believe not. I think this is a counterexample: Let $A$ be free on generators $x, y, z$ with $x=0, y = 1, z = 2$ and $d(y) = xz$, while let $B$ be the quotient DGA $k[x,z] / (xz)$. The quotient map from $A$ to $B$ is a quasiiso. If you push out along the map $k[x] \to k$, then you get a map $k[z] \otimes \Lambda[y] \to k[z]$ which is not a quasiiso. $\endgroup$ – Tyler Lawson Feb 20 '17 at 21:55

$\begingroup$ True, thanks. Might the model structures on dgcoalgebras or dgLie algebras in nonnegative/positive degrees be proper? (Sorry for the randomseeming questions, I just happen to be in need for a proper model structure for classical rational homotopy theory.) $\endgroup$ – Urs Schreiber Feb 21 '17 at 8:02

$\begingroup$ @Urs I'm afraid that I don't know in those circumstances; those model structures are a little outside my wheelhouse. I'd strongly suspect that the model structure on dgLie algebras in nonnegative degrees is left proper by a similar argument to the one here, since the condition that "noninjective in degree 0 can still be a cofibration" is replaced with "nonsurjective in degree 0 can still be a fibration". $\endgroup$ – Tyler Lawson Feb 22 '17 at 5:22
If you want a reference other than ToenVezzosi, 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 BataninBerger Homotopy theory for algebras over polynomial monads, which checks that chain complexes in characteristic zero are hmonoidal 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 BataninBerger paper.