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Acyclic models

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Suppose $R$ is a ring. Let $\mathcal{Mod}$ (or $R-\mathcal{Mod}$)be the category of modules over $R$ and $Ch(\mathcal{Mod})$ be the category of chain complexes of $R$-modules.

Free functors

Let $\mathcal{K}$ be a category with models $\mathcal{M}$, which is a set objects of $\mathcal{K}$: $$\mathcal{M} \subset \text{Obj}(\mathcal{K}).$$

We say that functor $\mathcal{F}: \mathcal{K} \to \mathcal{Mod}$ is free with basis in $\mathcal{M}$ if for each $K \in \text{Obj}(\mathcal{K})$, $\mathcal{F}(K)$ is a free module with basis $$Q_K=\{ \mathcal{F} (\sigma )(N): N =\mathcal{F}(M), M \in \mathcal{M}, \sigma \in \text{Hom}(M,K) \}.$$

The set $\mathcal{X} = F(\mathcal{M})$ is called a $\mathcal{F}$-model set.

Example a free functor with basis:

  • $\mathcal{K} = \mathcal{Top}$;
  • $\mathcal{F}=S_k$ is the functor of the $k$-th term of the singular chain complex;
  • $\mathcal{M}=\{\Delta ^k\}$, where $\Delta ^k =[e_0,...,e_k]$ is the standard $k$-simplex, so $\mathcal{X}=\{\text{id}_{\Delta ^k}\}$.

Indeed, for any topological space $X$ $$Q_X=\{ S_k (\sigma )(\text{id}_{\Delta ^k}): \sigma \in \text{Hom}(\Delta ^k,K) \}$$ is a basis of $S_k(X)$.

Non-example a free functor with basis:

  • $\mathcal{K}$ = simplicial complexes;
  • $\mathcal{F}=S_k$ is the functor of the $k$-th term of the simplicial chain complex;
  • $\mathcal{M}=\{\Delta ^k\}$, so $\mathcal{X}=\{ \Delta ^k \}$.

Indeed, for a simplicial complex $K$ $$Q_K=\{ C_k (\sigma )(\Delta ^k): \sigma \in \text{Hom}(\Delta ^k,X) \}$$ is not a basis of $C_k(K)$ because $Q_K$ contains each $k$-simplex $\Delta ^k$ of $K$ along with a simplex produced by every possible permutation $s$ of its vertices: $\Delta ^k_s =[s(e_0),...,s(e_k)]$.

Theorem. Suppose $\mathcal{F}: \mathcal{K} \to \mathcal{Mod}$ is a free functor with basis in models $\mathcal{M}$. If $\mathcal{G}: \mathcal{K} \to \mathcal{Mod}$ is another functor, then there is a unique natural transformation $\tau : \mathcal{F} \to \mathcal{G}$ with $$\tau _M(\mathcal{F}(M))=\mathcal{G}(M),\forall M \in \mathcal{M}.$$

It is illustrated by the diagram: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{cccccccccc} {\mathscr F}(M) & \ra{\tau _M} & {\mathscr G}(M) \\ \da{ {\mathscr F}(\sigma) } & & \da{ {\mathscr G}(\sigma) } \\ {\mathscr F}(K) & \ra{\tau _M} & {\mathscr G}(K) \end{array} $$ for $\sigma \in \text{Hom}(M,K)$.

Proof. Rotman p. 241. $\blacksquare$

Theorem. Suppose $\mathcal{F}: \mathcal{K} \to \mathcal{Mod}$ is a free functor with basis in models $\mathcal{M}$. Suppose the second square of the following diagram of functors and natural transformations is commutative (for every object $C$): $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{cccccccccc} {\mathscr F} & \ra{\tau} & {\mathscr G} & \ra{\mu} & {\mathscr H}\\ ?\da{\gamma} & & \da{ \beta } & & \da{ \alpha}\\ {\mathscr F'} & \ra{\tau '} & {\mathscr G'} & \ra{\mu '} & {\mathscr H'}. \end{array} $$ Suppose

  • the top row is zero: $\mu _C \tau _C =0$ for every object $C$;
  • the bottom row is exact: $\text{im} \tau '_M = \ker \mu '_M$ for every model $M$.

Then there is a natural transformation $\gamma :\mathscr{F} \to \mathscr{F}'$ that makes the first square commute.

Proof. Rotman p. 241. $\blacksquare$

Acyclic models

We will consider positive chain complexes: $C_i = 0$ for $i < 0$.

Observe that every positive chain complex $(C,\partial)$: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccc} ... & \ra{} & C_{k+1} & \ra{\partial_{k+1}} & C_{k} & \ra{\partial_{k}} & C_{k-1} & \ra{\partial_{k-1}} & C_{k-2} & \ra{} & ...& \ra{\partial_{2}} & C_{1} & \ra{\partial_{1}} & C_{0} \end{array} $$ can be extended to $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccc} ... & \ra{} & C_{k+1} & \ra{\partial_{k+1}} & S_{k} & \ra{\partial_{k}} & C_{k-1} & \ra{\partial_{k-1}} & C_{k-2} & \ra{} & ... & \ra{\partial_{2}} & C_{1} & \ra{\partial_{1}} & C_{0} & \ra{\epsilon} & H_0(C) & \ra{} & 0, \end{array} $$ because $H_0(C)=\text{coker}\partial _1$.

Theorem. Suppose

  • $(F,d),(G,\partial)$ are two positive chain complexes,
  • $F$ is free, and
  • $G$ is acyclic. Then for any maps $e,\epsilon, \varphi$:

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccc} F_{0} & & G_0\\ \da{e} & & \da{\epsilon} \\ H_0(F) & \ra{\varphi} & H_0(G),\\ \end{array} $$ there is a chain map $f:F\to E$ that make this diagram commutative: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccc} \da{d_3} & & \da{\partial_3} \\ F_2 & \ra{f_2} & G_2\\ \da{d_2} & & \da{\partial_2} \\ F_{1} & \ra{f_{1}} & G_1\\ \da{d_1} & & \da{\partial_1} \\ F_{0} & \ra{f_{0}} & G_0\\ \da{e} & & \da{\epsilon} \\ H_0(F) & \ra{\varphi} & H_0(G),\\ \da{} & & \da{} \\ 0& &0\\ \end{array} $$ Then we say that $f$ is a chain map over $\varphi$.

Proof. Induction. Rotman p. 238. $\blacksquare$

Theorem. Let $$\mathcal{F},\mathcal{G} : \mathcal{K} \to Ch(\mathcal{Mod})$$ be positive functors and $$\mathcal{F}_i,\mathcal{G}_i : \mathcal{K} \to \mathcal{Mod}$$ are the corresponding "coordinate" functors. Suppose

  1. $\mathcal{F}_k$ is a free functor with basis in models $\mathcal{M}_k$ for all $k \ge 0$;
  2. $\mathcal{G}$ is acyclic at the models, i.e., $H_k(\mathcal{G}(M)) = 0$ for all $k>0$ and all $M \in \mathcal{M}_k$.

Then:

  • (1) For every natural transformation $\varphi : H_0(\mathcal{F}) \to H_0(\mathcal{G})$ there is a natural chain map $f : \mathcal{F} \to \mathcal{G}$ over $\varphi$:

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccc} \da{d_3} & & \da{\partial_3} \\ \mathcal{F}_2(K) & \ra{f_2} & \mathcal{G}_2(K)\\ \da{d_2} & & \da{\partial_2} \\ \mathcal{F}_{1}(K) & \ra{f_{1}} & \mathcal{G}_1(K)\\ \da{d_1} & & \da{\partial_1} \\ \mathcal{F}_{0}(K) & \ra{f_{0}} & \mathcal{G}_0(K)\\ \da{\epsilon} & & \da{e} \\ H_0(\mathcal{F}(K)) & \ra{\varphi} & H_0(\mathcal{G}(K))\\ \da{} & & \da{} \\ 0& &0\\ \end{array} $$

  • (2) Suppose in addition $\mathcal{G}$ satisfies condition 1 above and $\mathcal{F}$ condition 2. Then, if $f,g: \mathcal{F} \to \mathcal{G}$ are natural chain maps over $\varphi$, then they are naturally chain homotopic.

Proof. Follows from the previous theorem. Rotman p. 243. $\blacksquare$

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