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\section{The method of acyclic models}  \label{sec:moam}

Let $F_*$ and $G_*$ be chain complexes.

Assume $F_k$ has a basis $\{x_{kj}\}$

(that is, $F_*$ is free and we have specified a basis).

(In our applications, $\{x_{kj}\}$ will typically be singular $k$-simplices or 

$k$-blob diagrams.)

For each basis element $x_{kj}$ assume we have specified a ``target" $D^{kj}_*\sub G_*$.

We say that a chain map $f:F_*\to G_*$ is {\it compatible} with the above data (basis and targets)

if $f(x_{kj})\in D^{kj}_*$ for all $k$ and $j$.

Let $\Compat(D^\bullet_*)$ denote the subcomplex of maps from $F_*$ to $G_*$

such that the image of each higher homotopy applied to $x_{kj}$ lies in $D^{kj}_*$.

\begin{thm}[Acyclic models]  \label{moam-thm}

Suppose 

\begin{itemize}

\item $D^{k-1,l}_* \sub D^{kj}_*$ whenever $x_{k-1,l}$ occurs in $\bd x_{kj}$

with non-zero coefficient;

\item $D^{0j}_0$ is non-empty for all $j$; and

\item $D^{kj}_*$ is $(k{-}1)$-acyclic (i.e.\ $H_{k-1}(D^{kj}_*) = 0$) for all $k,j$ .

\end{itemize}

Then $\Compat(D^\bullet_*)$ is non-empty.

If, in addition,

\begin{itemize}

\item $D^{kj}_*$ is $m$-acyclic for $k\le m \le k+i$ and for all $k,j$,

\end{itemize}

then $\Compat(D^\bullet_*)$ is $i$-connected.

\end{thm}

\begin{proof}

(Sketch)

We will build a chain map $f\in \Compat(D^\bullet_*)$ inductively.

Choose $f(x_{0j})\in D^{0j}_0$ for all $j$

(possible since $D^{0j}_0$ is non-empty).

Choose $f(x_{1j})\in D^{1j}_1$ such that $\bd f(x_{1j}) = f(\bd x_{1j})$

(possible since $D^{0l}_* \sub D^{1j}_*$ for each $x_{0l}$ in $\bd x_{1j}$

and $D^{1j}_*$ is 0-acyclic).

Continue in this way, choosing $f(x_{kj})\in D^{kj}_k$ such that $\bd f(x_{kj}) = f(\bd x_{kj})$

We have now constructed $f\in \Compat(D^\bullet_*)$, proving the first claim of the theorem.

Now suppose that $D^{kj}_*$ is $k$-acyclic for all $k$ and $j$.

Let $f$ and $f'$ be two chain maps (0-chains) in $\Compat(D^\bullet_*)$.

Using a technique similar to above we can construct a homotopy (1-chain) in $\Compat(D^\bullet_*)$

between $f$ and $f'$.

Thus $\Compat(D^\bullet_*)$ is 0-connected.

Similarly, if $D^{kj}_*$ is $(k{+}i)$-acyclic then we can show that $\Compat(D^\bullet_*)$ is $i$-connected.

\end{proof}

\nn{do we also need some version of ``backwards" acyclic models?  probably}