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Sep 1, 2020 at 7:37 comment added Urs Schreiber Hm, isn't it Lemma 14.2 in Felix-Halperin-Thomas?
Sep 1, 2020 at 6:46 comment added Urs Schreiber edit: Manetti-Meazzini talk about co-connective, not connective dgca-s
Sep 1, 2020 at 6:37 comment added Urs Schreiber Hm, Corollary 3.4 in Manetti-Meazzini arxiv.org/abs/1802.06707v1 says that connective dgc-algebras is left proper, after all. (I guess in the above counter-example, the morphism pushed out along was not a cofibration.)
Feb 22, 2017 at 5:22 comment added Tyler Lawson @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 dg-Lie algebras in nonnegative degrees is left proper by a similar argument to the one here, since the condition that "non-injective in degree 0 can still be a cofibration" is replaced with "non-surjective in degree 0 can still be a fibration".
Feb 21, 2017 at 8:02 comment added Urs Schreiber True, thanks. Might the model structures on dg-coalgebras or dg-Lie algebras in non-negative/positive degrees be proper? (Sorry for the random-seeming questions, I just happen to be in need for a proper model structure for classical rational homotopy theory.)
Feb 20, 2017 at 21:55 comment added Tyler Lawson @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 quasi-iso. 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 quasi-iso.
Feb 20, 2017 at 18:29 comment added Urs Schreiber This is for unbounded dgca-s, right? Does this also go through for the category of cochain dgc-algebras in non-negative 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.
Jun 16, 2015 at 13:25 comment added Dmitri Pavlov 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.
May 2, 2015 at 18:19 vote accept David Carchedi
May 1, 2015 at 5:17 history answered Tyler Lawson CC BY-SA 3.0