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Post Undeleted by Joey Hirsh
Post Deleted by Joey Hirsh
2 added 12 characters in body

I think the answer to question 1 is "Yes." Here's why:

Fix a field k of characteristic 0 (you asked about k = Q).

Let C and D be (graded-)commutative DG k-algebras. Then there exists a noncommutative quasi-isomorphism from C to D iff there exists an $A_{\infty}$ map Quasi-iso $F:C \to D$. (In fact more is true---there is a one-to-one correspondence between equivalence classes of noncommutative quasi-iso's and equivalence classes of $A_\infty$ morphisms.)

Since there exists a commutative quasi-isomorphism from C to D iff there exists a $C_{\infty}$ map quasi-iso $G: C \to D$, your question can be rephrased as follows: "Suppose C,D are commutative DG k-algebras, and $F: C \to D$ is an $A_{\infty}$ map. Does this imply the existence of a $C_{\infty}$ map?"

We can describe $A_{\infty}, C_{\infty}$ maps as follows: for a dga A, we can consider the cofree (conilpotent) coalgebra $T^c(A)$, and we observe that the dga structure maps $d: A \to A, \cdot: A \otimes A \to A$ define a map $T^c(A) \to A$ by projecting onto $A \oplus (A \otimes A)$ and then doing $d$ on the first factor and $\cdot$ on the second. This map can be extended as a coderivation $D_A:T^c(A) \to T^c(A)$, and the claim is that $D_A^2 = 0.$ An $A_{\infty}$ map $F:A \to B$ is by a definition a dg-coalgebra map $F:(T^c(A), D_A) \to (T^c(B), D_B)$. Similarly, a commutative dga gives rise to a cofree cocommutative dg coalgebra, using $S^c$, and $C_{\infty}$ morphisms are dg-coalgebra maps between these. An $C_{\infty}$ map is actually an $A_{\infty}$ map with some symmetry properties (since $S^c \hookrightarrow T^c$).

Here's a first step in the argument "up to signs" that you can symmetrize the $A_{\infty}$ map F.

To get a $C_{\infty}$ map from F, write $F = f + f_2 + f_3 + \cdots$, where $f_n : C^{\otimes n} \to D$. Then the condition that $F$ is an $A_{\infty}$ morphism gives $(df_2)(x,y) = f(x) \cdot_{D} f(y) \pm f(x \cdot_{C} y)$ where $df_2$ should be understood as the differential in the Hom complex $Hom(C^{\otimes 2}, D)$. Since the products in C and D are commutative, we actually get: $(df_2)(x,y) = f(x) \cdot_{D} f(y) \pm f(x \cdot_{C} y) = \pm f(y) \cdot_D f(x) \pm f(y \cdot_C x) = (df_2)(y,x)$ so we see that at least $(df_2)$ is graded symmetric. Since k has characteristic 0, we can define

$g_2 := \frac{1}{2} \left( f_2 + f_2^{op} \right)$

and since

$(dg_2) = \frac{1}{2} d \left( f_2 + f_2^{op} \right) = \frac{1}{2} \left( df_2 + d(f_2^{op}) = \frac{1}{2}( df_2 \pm (df_2)^{op} ) = (df_2) \right)$

we see that $dg_2$ still provides the homotopy between $f( \cdot_C )$ and $f( ) \cdot f( )$.

I think an argument along these lines will also provide suitable $g_n$ for $n \geq 3$.

I wonder if someone can provide an "abstract nonsense" answer along the following lines: the functor i: Commutative DGAs -----> nonCommutative DGAs has a left adjoint which preserves quasi-iso's (does abelianization preserve maps inducing isomorphisms in Homology?), and so we can verify that the image of Commutative DGAs in (nonCommutative DGAs)$[W^{-1}]$ has the universal property that a localization (Commutative DGAs)$[W^{-1}]$ should have... (Where $W$ above should be taken to be the collection of maps inducing iso's in homology in each category).

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