Let $\mathscr{C}$ be an abelian category. If $P_\bullet \in \operatorname{Ch}_{\geq 0}(\mathscr{C})$ is a bounded below complex of projectives, and $C_\bullet \in \operatorname{Ch}_{\geq 0}(\mathscr{C})$ is a bounded below exact complex, then $[P_\bullet, C_\bullet] = 0$. (Every chain map $P_\bullet \to C_\bullet$ is nullhomotopic.)

It is tempting to conjecture that

If $\mathscr{B} \underset{G}{\overset{F}{\rightrightarrows}} \operatorname{Ch}_{\geq 0}(\mathscr{C})$ are functors from an arbitrary category $\mathscr{B}$, where $F$ lands in projective complexes, and $G$ lands in acyclic complexes, then $[F,G]=0$, meaning that any natural transformation from $F \Rightarrow G$ is naturally chain homotopic to the zero natural transformation.

The acyclic models theorem implies something similar: that if $\mathscr{B}$ has models $\mathcal{M}$, and $F$ is a free functor w.r.t. $\mathcal{M}$, and if $G$ is acyclic, then $[F,G]$ is indeed zero.

Is the highlighted theorem above untrue?

Given a natural transformation, I can choose a map $\tau: (FX)_\bullet \to (GX)_\bullet[1]$ for each object $X \in \mathscr{B}$. Making $\tau$ natural is the problem. If I try to define the first map $\tau_0$ in the natural transformation $\tau$, and check whether it is natural, I find that the naturality diagram \begin{array}{} FX_0 & \xrightarrow{(Ff)_0} &FY_0\\ (\tau_X)_0 \downarrow && \downarrow (\tau_Y)_0 \\ GX_1 & \xrightarrow{(Gf)_1} & GY_1 \end{array} only commutes up to a boundary element in $\partial^{GY}(GY_2)$.

  • $\begingroup$ More generally, we have $[P_\bullet,C_\bullet]=0$ for $P_\bullet, C_\bullet\in\text{Ch}(\mathscr{C})$, with $C_\bullet$ acyclic, $P_n$ projective for all $n$, and at least one of $P_\bullet$ or $C_\bullet$ is in $\text{Ch}_{\geq 0}(\mathscr{C})$. (I was wondering myself after I read the first two lines of the post.) $\endgroup$ May 22, 2023 at 6:46

2 Answers 2


The trouble is that an object of the functor category $[\cal B,\cal C]$ need not be projective even if every object in the image is. I think what you would get would be a weak homotopy (homotopy at each object of $\cal B$). For more on this subject, we my book titled Acyclic Models, which deals with every version of the theorem I am aware of.

  • $\begingroup$ Thanks for your reply. I have a couple of questions about what you said. First, if we go back to chain complexes, the projectivity of the object $P_\bullet \in \operatorname{Ch}_\bullet(\mathscr{C})$ is not what interests us there either, right? Secondly, is there a characterization of projective objects in $\mathsf{Fun}(\mathscr{B}, \operatorname{Ch}_{\geq 0}(\mathscr{C}))$? $\endgroup$
    – Eric Auld
    Jul 10, 2016 at 18:52
  • $\begingroup$ I don't know of any, although I haven't thought about. $\endgroup$ Jul 11, 2016 at 12:08

I also came up with this question recently and made some try.

As far as I am concerned, the correct generalization of "free with basis in $\mathcal{M}$" is "projective with basis in $\mathcal{M}$".

Definition: A functor $S: \mathcal{C}\longrightarrow Mod_{R}$ is said to be projective with basis in $\mathcal{M}$ if the following two conditions hold

  1. $T(C)$ is projective for all $C\in\mathcal{C}$.
  2. There is a $T$-model set $\chi=\{x_\lambda\in T(M_\lambda)\mid M_\lambda\in \Lambda\}$ s.t. $$ \{T(g)(x_\lambda)|g\in \hom(M_\lambda,C), \lambda\in \Lambda\} $$ is projective basis for $T(C)$. i.e. For each $x\in T(C)$, it can be expressed as $$ x=\sum_{\lambda\in \Lambda}\sum_{g\in \hom(M_\lambda,C)} f_{g,\lambda}^C(x)T(g)(x_\lambda) $$ where $\{f_{q,\lambda}^C: T(C)\longrightarrow R\}$ is a fixed set of morphisms of $R$-modules.

A functor $S_\bullet:\mathcal{C}\longrightarrow Comp_R$, where $Comp_R$ is the category of chain complex of $R$-modules, is said to be projective with basis in $\mathcal{M}$ if each $S_n$ is projective with basis in $\mathcal{M}$.

And we can state a proposition:

Proposition: Suppose $\mathcal{C}$ is a category with models $\mathcal{M}$. Suppose $T_\bullet, S_\bullet:\mathcal{C}\longrightarrow Comp_R$ are two functors such that both $T_\bullet$ and $S_\bullet$ are non-negative. Assume further $T_\bullet$ is projective with basis in $\mathcal{M}$ and $S_\bullet$ is acyclic in the positive degree on each element $M\in\mathcal{M}$. Suppose $$ \Theta: H_0\circ T_\bullet\longrightarrow H_0\circ S_\bullet $$ is a natural transformation. $\exists $ a natural chain morphism $\Psi_\bullet:T_\bullet\longrightarrow S_\bullet$ which is unique up to natural chain homotopy and has $H_0(\Psi_\bullet)=\Theta$.

And this proposition seems to be a specialization of Theorem 1 in

Dold A., MacLane S., Oberst U. (1967) Projective classes and acyclic models. In: Reports of the Midwest Category Seminar. Lecture Notes in Mathematics, vol 47. Springer, Berlin, Heidelberg

Hope that helps.


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