Let $\Delta^n$ be the standard $n$-simplex, that is, $\Delta^n = {(x_0, x_1, \ldots, x_n) \in \mathbb{R}^{n+1} : x_i \geq 0, \sum_{i=0}^n x_i = 1}$. We will use the barycentric coordinates for the points in $\Delta^n$, which are defined as follows: for $0 \leq i \leq n$, let $e_i \in \Delta^n$ be the point with $x_i = 1$ and $x_j = 0$ for $j \neq i$. Then any point $p \in \Delta^n$ can be written uniquely as a convex combination of the $e_i$’s, that is, $p = \sum_{i=0}^n \lambda_i e_i$, where $\lambda_i \geq 0$ and $\sum_{i=0}^n \lambda_i = 1$.

Now let $\phi: [m] \to [n]$ be an order-preserving function, and let $\phi:\Delta^m \to \Delta^n$ be the affine map that extends $\phi$ to the standard simplices. We want to express $\phi(p)$ in terms of the barycentric coordinates of $p = \sum_{i=0}^m \lambda_i e_i$.

Note that $\phi(e_i)$ is a point in $\Delta^n$, so we can write $\phi(e_i)$ as a convex combination of the $e_j$’s. Let $\phi(e_i) = \sum_{j=0}^n a_{ij} e_j$, where $a_{ij} \geq 0$ and $\sum_{j=0}^n a_{ij} = 1$. Then for any $p \in \Delta^m$, we have \begin{align*} \phi(p) &= \phi\left(\sum_{i=0}^m \lambda_i e_i\right) \ &= \sum_{i=0}^m \lambda_i \phi(e_i) \ &= \sum_{i=0}^m \lambda_i \sum_{j=0}^n a_{ij} e_j \ &= \sum_{j=0}^n \left(\sum_{i=0}^m \lambda_i a_{ij}\right) e_j. \end{align*}

Let $t_j = \sum_{i=0}^m \lambda_i a_{ij}$, so we have $\phi(p) = \sum_{j=0}^n t_j e_j$. Since $e_j$ is the point with $x_j = 1$ and $x_i = 0$ for $i \neq j$, we have $t_j = \phi(\lambda_j)$, where $\lambda_j$ is the $j$th barycentric coordinate of $p$. Therefore, we have

$t_j=\sum_{i=0}^m \lambda_i a_{i j}$,

where $a_{ij}$ is the $j$th barycentric coordinate of $\phi(e_i)$.

Let $X$ be a topological space. We denote by $\pi_0(X)$ the set of path-components of $X$. That is, for any two points $x,y\in X$, we say that $x$ and $y$ are in the same path-component if there exists a path $\gamma:[0,1]\to X$ such that $\gamma(0)=x$ and $\gamma(1)=y$. The set $\pi_0(X)$ is a partition of $X$ into disjoint subsets, and each subset is a path-connected subspace of $X$.

Now, we construct an isomorphism between $\mathbb{Z} \pi_0(X)$ and $H_0(X)$. Let $\mathbb{Z}\pi_0(X)$ be the free abelian group generated by the set $\pi_0(X)$, and let $\phi:\mathbb{Z}\pi_0(X)\to H_0(X)$ be the homomorphism defined as follows. For each path-component $[x]\in \pi_0(X)$, choose a point $x\in [x]$, and define $\phi([x])=[x]$ to be the class in $H_0(X)$ represented by the point $x$. We extend $\phi$ to all of $\mathbb{Z}\pi_0(X)$ by linearity.

We claim that $\phi$ is an isomorphism. To see this, we construct its inverse. Let $\psi:H_0(X)\to \mathbb{Z}\pi_0(X)$ be the homomorphism defined by sending each class $[x]\in H_0(X)$ to the generator $[x]\in \mathbb{Z}\pi_0(X)$ corresponding to the path-component containing $x$. That is, if $[x]\in H_0(X)$, then there exists a chain $\sum_i n_i \sigma_i$ with $\sigma_i:\Delta^0\to X$ such that $\partial (\sum_i n_i \sigma_i)=[x]$, and we define $\psi([x])=[\sigma_i(0)]\in \mathbb{Z}\pi_0(X)$. Again, we extend $\psi$ to all of $H_0(X)$ by linearity.

It is easy to see that $\psi$ is well-defined and that $\phi$ and $\psi$ are inverses of each other, so $\phi$ is an isomorphism.

First, let’s assume that the chain complex $A$ is contractible. This means that there exists a chain map $s:A\rightarrow A$ (called a homotopy) such that $ds+sd=\mathrm{id}_A$, where $d$ is the differential of $A$. We can then define a chain map $r:A\rightarrow A$ by $r_n = \frac{1}{2}(1+s)$, which is well-defined because $s$ is a chain map. It’s easy to check that $r$ is also a chain map, and that $sr=\mathrm{id}_A$. This shows that $A$ is chain-homotopy-equivalent to the trivial chain complex.

Next, let’s assume that $A$ is acyclic and that the inclusion $Z_n A\hookrightarrow A_n$ is a split monomorphism for every $n$. Since $A$ is acyclic, we have $H_n(A)=0$ for all $n$. It remains to show that $A$ is chain-homotopy-equivalent to the trivial chain complex.

For each $n$, let $s_n:A_n\rightarrow A_{n+1}$ be a map such that $ds_n+s_{n-1}d=\mathrm{id}{A_n}$ (recall that we set $s{-1}=0$). Such a map exists because the inclusion $Z_n A\hookrightarrow A_n$ is a split monomorphism. In fact, we can take $s_n$ to be the composite of the projection $A_{n+1}\rightarrow Z_{n+1}A$ (which exists because of the split monomorphism) and the inclusion $Z_{n+1}A\hookrightarrow A_{n+1}$.

We claim that $s={s_n}$ is a chain homotopy between $A$ and the trivial chain complex. To see this, we need to show that $ds+sd=\mathrm{id}A$. We have: \begin{align*} (ds+sd)n &= d_ns{n-1}+s{n-1}d_n \ &= (s_{n-1}d-d_{n-1}s_n)+s_{n-1}d_n \ &= s_{n-1}(d_n+d_{n-1}) \ &= s_{n-1}^2 \ &= \frac{1}{4}(1+s_{n-1})^2 \ &= \frac{1}{4}(1+2s_{n-1}+s_{n-1}^2) \ &= \frac{1}{4}(1+2s_{n-1}+(1+s_{n-2})s_n) \ &= \frac{1}{4}(2s_n+s_{n-1}s_n+s_{n-2}s_n) \ &= \frac{1}{4}(2s_n+s_{n-1}d_n+s_{n-2}d_{n-1}) \ &= \frac{1}{4}(2s_n+d_{n-1}s_n+d_ns_{n-1}) \ &= \frac{1}{4}(2s_n+d_ns_{n-1}+s_{n-1}d_n) \ &= \frac{1}{4}(2s_n+\mathrm{id}{A_n}) \ &= r_n, \end{align*}