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

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

    39 \nn{...}

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

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

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

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

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

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

    45 

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

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

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

    49 between $f$ and $f'$.

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

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