The cover-small inclusion is a chain homotopy equivalence

Use $S$ and $T$ with $1-S=\mathrm{\partial }T+T\mathrm{\partial }$ from Subdivision is chain homotopic to the identity. Their constructions in Barycentric subdivision operator and Subdivision prism homotopy take each singular simplex to a finite sum of its compositions with affine simplices in its own domain. Thus neither operator enlarges the image of a simplex, and both preserve $A_{\bullet }$. Also $S$ is a chain map: applying $\mathrm{\partial }$ on either side of the displayed homotopy identity gives $\mathrm{\partial }S=S\mathrm{\partial }$. Set $D_q={\sum }_{i=0}^{q-1}T{S}^i$ for $q\ge 0$, so $D_0=0$ and $1-S^q=\mathrm{\partial }{D}_q+D_q\mathrm{\partial }$ by telescoping.

For each singular simplex $\sigma$, let $a(\sigma )$ be the least $q\ge 0$ with $S^q\sigma \in A_{\bullet }$; existence follows from Finite chains eventually become cover-small with integer coefficients. Define $m$ recursively on dimension: on vertices set $m=0$, and for positive-dimensional $\sigma$ set $m(\sigma )=\max (\{a(\sigma )\}\cup \{m(\sigma \circ {\delta }_j):0\le j\le \dim \sigma \})\mathrm{.}$ Every maximum is finite and uses already defined lower-dimensional values. Since $S$ preserves small chains, $S^{m(\sigma )}\sigma$ is small. Every face $\tau$ has $m(\tau )\le m(\sigma )$ by construction, regardless of cancellations in subdivided chains. If $\sigma$ is already small, all its faces are small, so this recursion gives $m(\sigma )=0$.

Define $D\sigma =D_{m(\sigma )}\sigma$ on integral generators and extend linearly. Put $R=1-\mathrm{\partial }D-D\mathrm{\partial }$ on $C_{\bullet }$. The identity ${\mathrm{\partial }}^2=0$ gives $\mathrm{\partial }R=R\mathrm{\partial }$. On a simplex, telescoping gives $R\sigma =S^{m(\sigma )}\sigma +D_{m(\sigma )}(\mathrm{\partial }\sigma )-D(\mathrm{\partial }\sigma )\mathrm{.}$ The first term is small. For each face $\tau$, its signed correction is ${\sum }_{i=m(\tau )}^{m(\sigma )-1}T{S}^i\tau$. Each $S^i\tau$ is small for these indices, and $T$ preserves small chains. Therefore $R\sigma$ is small.

Regard $R$ as a chain map $r:C_{\bullet }\to A_{\bullet }$ and let $\iota :A_{\bullet }\hookrightarrow C_{\bullet }$. Then $1-\iota r=\mathrm{\partial }D+D\mathrm{\partial }$. On every small simplex $m=0$, hence $D=0$ on $A_{\bullet }$; its boundary is also small, so $r\iota =1$. Thus one inverse composite is the identity and the other is chain homotopic to it. In degree zero all vertices are small and $D=0$; if $X$ is empty both complexes are zero.

The group $A_n$ is the direct summand of the free abelian group $C_n$ spanned by small singular simplices. Tensoring its inclusion with $G$ identifies $A_n\otimes G$ with the cover-small subgroup of $C_n(X;G)$. Tensor $r,D$ and their identities with ${\mathrm{id}}_G$; these identities remain valid for every abelian $G$, including $G=0$, without a flatness assumption. This proves the stated chain homotopy equivalence with the page's coefficients.