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@\subsection{Hyperchain}
We assume some class of topological chains on $X$ is given, i.e. semialgebraic chains. For any open set $U\subset X$ we denote by $C_i(U)$ the group of chains with support on $U$ of dimension $i$. Suppose an abstract cell complex $\sigma$ and a hypercover $(U_a)$ is given. 

The complex of \emph{hyperchains} is defined similarly to the complex of hypersections but with change of sign. We put
\[
C_\sigma(X)_i = \bigoplus_{j\geq 0} C_{\sigma_j}(X)_{i-j}, \qquad\text{where}
\]
\[
C_{\sigma_j}(X)_{i-j} = \bigoplus_{a\in \sigma_j} C_{i-j}(U_a).
\]
Given a hyperchain $\xi\in C_{i-j}(U_a)\subset C_\sigma(X)_i$ its boundary is 
\[
\partial_\sigma \xi:= \partial\xi - (-1)^i \sum_{b\in\sigma_{j-1}} D_{a b} \xi_b.
\]
Here $\partial\xi\in C_{i-1-j}(U_a)$ is the ordinary boundary of $\xi$ and $\xi_b$ denotes the same chain as $\xi$, but considered as an element of $C_{i-j}(U_b)$ if $U_a\subset U_b$. The definition of the boundary map is then extended to $C_\sigma(X)_i$ by linearity. 

The augmentation morphism $\epsilon:C_\sigma(X)\rightarrow C(X)$ sends all chains in $C_{i-j}(U_a)$ with $j>0$ to $0$ and the ones with $j=0$ to themselves.

We have the following lifting property:
\begin{prop}
Let $\xi$ be a chain on $X$, $\eta=\partial \xi$, $\bar\eta$ a hyperchain such that $\epsilon\bar\eta=\eta$ and $\partial_\sigma \bar\eta=0$. Then there exists a hyperchain $\bar\xi$ such that $\partial_\sigma\bar\xi=\bar\eta$ and $\epsilon \bar\xi=\xi$.
\end{prop}
\begin{proof}
The complex $C_\sigma(X)_\bullet$ is the total complex of the bicomplex $C_{\sigma_\bullet}(X)_\bullet$ with the horizontal differentials induced by the boundary maps of chains in the space $X$ and the vertical ones induced by the boundary maps of the complex $\sigma$:
\[
\begin{CD}
& & C_{\sigma_1}(X)_{i-1} @@>>> C_{\sigma_1}(X)_{i-2} \\
& & @@VVV @@VVV\\
C_{\sigma_0}(X)_i @@>>> C_{\sigma_0}(X)_{i-1} @@>>> C_{\sigma_0}(X)_{i-2}\\
@@V{\epsilon}VV @@V{\epsilon}VV\\
C(X)_i @@>>> C(X)_{i-1}
\end{CD}
\]
Clearly it is enough to prove that the vertical complexes are exact.

Each group $C_{\sigma_j}(X)_i$ is a free abelian group generated by certain types of simplices in open subsets of $X$. For each simplex $s$ in $X$ there is a corresponding direct summand of $C_{\sigma_j}(X)$ which is generated by simplices in $U_a$ which coincide with $s$ in $X$. This makes a decomposition of $C_{\sigma_j}(X)_i$ into a direct sum. The decomposition clearly commutes with are so is decomposable into a direct sum of abeliancopies of $Z$ indexed by simplices in $X$. and the vertical differentials 
\end{proof}


Let the dimension of $\xi$ be $i$. Let $\bar\eta=\sum_{a\in\sigma} \eta_a$ with $\eta_a\in C_{i-1-\dim a}(U_a)$. We construct $\bar\xi=\sum_{a\in\sigma} \xi_a$ with $\xi_a\in C_{i-\dim a}(U_a)$. The conditions they need to satisfy are:
\[
\eta_b = \partial \xi_b - (-1)^i\sum_{a\in\sigma_j} D_{ab}\xi_a, \qquad b\in\sigma_{j-1};
\]
\[
\xi = \sum_{a\in \sigma_0} \xi_a.
\]
On the step $0$ we simply split $\xi$ into pieces $\xi_a$, $a\in\sigma_0$ so that $\xi_a\in\C_i(U_a)$. This is possible since $U_a$ for $a\in\sigma_0$ cover $X$. Thus the second condition is satisfied. Then we choose $\xi_a$ for $a\in\sigma_1$ such that the first condition is satisfied for $j=1$. On step $k$ we construct $\xi_a$ for $a\in\sigma_k$ such that the first condition is satisfied for $j=k$. We prove that it is always possible.












Let us define for every $Z\in M$ and $L\subset\{1,\ldots,n\}$ a subvariety $Z_L\subset Z$. The definition is inductive. We put
\begin{align*}
Z_{\varnothing} &= Z,\\
Z_{L\cup \{k\}} &= Z_L \cdot \pi_k^{-1} S_k\;\text{for $\max L<k\leq n$},
\end{align*}
where $\cdot$ means the following operation: for every irreducible component $Y$ of $Z_L$ consider $Y'=Y\cap \pi_k^{-1} S_k$; take the union of those $Y'$ which are strictly smaller than $Y$.

It is clear that the subvarieties $Z_L$ satisfy the following properties:
\begin{prop}
For any $Z\in M$ and $L\subset\{1,\ldots,n\}$ the subvariety $Z_L$ is either empty or is the union of a finite number of varieties from $M$ of dimension $\dim Z - |L|$.
\end{prop}

The main property of good choices of numbers is the following:
\begin{prop}\label{prop:nhp}
If $\vec\veps=(\veps_1,\dots,\veps_n, \veps_1',\dots,\veps_n')$ is a good choice of numbers with the corresponding sequence $\delta_1<\cdots<\delta_n$, then for every $Z\in M$ and $L\subset\{1,\ldots,n\}$ the intersection $Z\cap \pi_L^{-1}R_L(\vec\veps)$ is contained in the $\delta_{\veps_{\max L}}$-neighbourhood of $Z_L$.
\end{prop}
\begin{proof}
The proof goes by induction on $|L|$ with the base of induction being obvious. Let $L\subset\{1,\ldots,n\}$ and $\max L<k\leq n$. Let $x$ be a point belonging to the intersection $Z\cap \pi_{L\cup\{k\}}^{-1}R_{L\cup\{k\}} (\vec\veps)$. By the assumption of induction 
$\dist(x, Z_L)<\delta_{\max L} \leq \delta_{k-1}$. Therefore there exists an irreducible component $Y$ of $Z_L$ such that $\dist(x, Y)<\delta_{k-1}$. On the other hand $\pi_k(x)\in R_k(\veps_k, \veps_k')$. This means $\veps_k'<\dist(\pi_k(x),S_k)<\veps_k$. Let $Y'=Y\cap \pi_k^{-1}S_k$. Since $Y\in M$ one has $\dist(x, Y')<\delta_k$ by the second property of good choices. If $Y'\neq Y$ then $Y$ is a subset of $Z_{L\cup\{k\}}$ and we are done. Otherwise we have $\dist(x, Y')=\dist(x,Y)<\delta_{k-1}$ and by the third property of good choices $\dist(\pi_k(x),S_k)\leq \veps_k'$, which is a contradiction.
\end{proof}

This immediately implies
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