Fix a colored operad, which I will leave implicit, and a field $\mathbb K$ of characteristic $0$. By algebra in this post I will mean a dg algebra over $\mathbb K$ for the given colored operad. I will let $Y$ denote any algebra, and I will let $X$ denote an algebra such that if you forget the differential, then $X$ is free on a well-ordered set of generators, and such that for each generator, its differential is a composition of strictly earlier (for the well-ordering) generators.

Here are some facts I know how to prove, but my proofs are in places long-winded. They are not in any way due to me — they are "classical results", where "classical" is defined as "something the author learned in graduate school". My question is: what is a good reference that proves these results?

  1. One may try to build an algebra homomorphism $\eta: X \to Y$ inductively. When trying to $\eta(x)$ for a generator $x$, the only condition is that $\partial \eta(x) = \eta(\partial x)$, the latter having already been defined. The induction can continue iff $\eta(\partial x)$ is exact (it is already closed), and this is measured by the class of $\eta(\partial x)$ in $\mathrm H_{\deg x - 1}(Y)$. Here $x$ has homological degree $x$, and I'm using homological conventions, so that $\deg(\partial x) = \deg x - 1$. I'm also being a bit sloppy with notation — really I mean $\mathrm H_\bullet($the part of $Y$ with the appropriate colors for $\eta(x)$ to be valued there$)$.

  2. Suppose that $\eta(\partial x)$ is exact. Different choices for $\eta(x)$ might affect whether later steps of the induction succeed. Changing $\eta(x)$ by something exact will not affect the success or failure of later steps. So the "true" set of choice for $\eta(x)$ is a torsor for $\mathrm H_{\deg x}(Y)$.

  3. Recall Sullivan's simplicial dg commutative algebra $\Omega(\Delta_\bullet)$, whose $k$-simplices are the dgca $\mathbb K[t_0,\dots,t_k,\partial t_0,\dots,\partial t_k] / (\sum t_i = 1, \sum \partial t_i = 0)$. By definition, the space $\hom_\bullet(X,Y)$ of homomorphisms $X \to Y$ is the simplicial set whose $k$-simplices are $\hom_{\text{algebras}}(X,Y \otimes \Omega(\Delta_k))$. This simplicial set satisfies the Kan horn filling condition.

  4. Suppose we have chosen a homomorphism $\eta : X \to Y$. What is the homotopy type of the connected component of $\eta$ in $\hom_\bullet(X,Y)$? For $k\geq 1$, the $k$th homotopy group $\pi_k(\hom_\bullet(X,Y);\eta)$ is an extension of abelian groups, one for each generator $x$ of $X$, such that the $x$th group describes $\pi_k($space of choices for $\eta(x))$. The $x$th group in the extension is precisely $\mathrm{H}_{\deg x + k}(Y)$.

  • $\begingroup$ One correction: I think my proof for 4 depends on earlier $\pi_k$s vanishing. I don't know whether 4 is true as stated. $\endgroup$ Feb 1, 2014 at 20:44

1 Answer 1


Theo, there is a nice, abstractly developed, obstruction theory in Baues's "Combinatorial Foundation of Homology and Homotopy":


Your context can be fitted into the general framework of this book with little effort. This would give you a much more structured obstruction theory than what you suggest in 1 and 2. Computations are possible, and Baues himself has successfully applied it in many papers (and other books).

In 3, that's going to be a Kan complex whenever $X$ is cofibrant, your conditions may not be enough, but probably close.

As for 4, and to some extent for a different answer to 1 and 2, you can take a look at Bousfield's "Homotopy spectral sequences and obstructions"


This is a long paper, and Bousfield gives usually very short conceptual arguments, so it's indeed very dense, not easy to understand, and very difficult for explicit computations, but it is an invaluable source of knowledge.

  • $\begingroup$ Great. For the colored operads I care about, the category of dg algebras admits a model structure in which fibrations are surjections and acyclics are quasiisomorphisms. In this case, everything is fibrant, and the conditions on $X$ imply that $X$ is cofibrant. I've always assumed that this is true for any colored operad, but I realize the proofs I've seen tend to take specific ones. $\endgroup$ Feb 1, 2014 at 20:47
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    $\begingroup$ Usually the only issue is in arity 0. $\endgroup$ Feb 1, 2014 at 23:31
  • $\begingroup$ Ah, OK. Actually, my operads are generated in arity 2, with quadratic relations in arity 3, and that's it. $\endgroup$ Feb 2, 2014 at 0:40

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