Singular Chains and Singular Homology

Let $t=(t_0,\dots ,t_{n-2})\in {\mathrm{\Delta }}^{n-2}$. By [L1], the composite ${\delta }_j{\delta }_i(t)$ is obtained by first inserting $0$ in slot $i$ and then inserting $0$ in slot $j$. Since $i<j$, the second insertion happens strictly to the right of the first one, so the resulting $(n+1)$-tuple has zeros exactly in positions $i$ and $j$.

Again by [L1], the composite ${\delta }_i{\delta }_{j-1}(t)$ first inserts $0$ in slot $j-1$ and then inserts $0$ in slot $i$. After the second insertion, the earlier zero has shifted one place to the right, so the final tuple also has zeros exactly in positions $i$ and $j$, with every other coordinate of $t$ appearing in the same relative order as in step 1.1. Therefore the two composites agree coordinatewise.

If $n=0$, then ${\mathrm{\partial }}_0=0$ by [L1], so the degree-zero boundary is already zero. If $n=1$ and $\sigma :{\mathrm{\Delta }}^1\to X$ is a singular $1$-simplex, then ${\mathrm{\partial }}_0{\mathrm{\partial }}_1\sigma ={\mathrm{\partial }}_0(\sigma {\delta }_0-\sigma {\delta }_1)=0,$ again because ${\mathrm{\partial }}_0=0$. Thus the claim holds in the two low degrees.

Assume $n\ge 2$ and let $\sigma :{\mathrm{\Delta }}^n\to X$ be a singular $n$-simplex. Expanding twice with [L1] gives ${\mathrm{\partial }}_{n-1}{\mathrm{\partial }}_n\sigma ={\sum }_{i=0}^n{\sum }_{j=0}^{n-1}(-1{)}^{i+j}\sigma {\delta }_i{\delta }_j\mathrm{.}$ Reindex the terms by pairs $0\le j<i\le n$ and $0\le i\le j\le n-1$. By [L2], the term $\sigma {\delta }_i{\delta }_j$ with $j<i$ equals $\sigma {\delta }_j{\delta }_{i-1}$, but the two appearances carry opposite signs because $(-1{)}^{i+j}=-(-1{)}^{j+(i-1)}\mathrm{.}$ Hence every codimension-two face occurs twice with opposite coefficients, so the whole sum is zero.

Since singular simplices generate $C_n(X;\mathbb{Z})$ and the coefficient version is obtained by tensor extension, steps 1.1 and 1.2 imply ${\mathrm{\partial }}_{n-1}{\mathrm{\partial }}_n=0$ on $C_n(X;G)$ for every $n\ge 1$, with degree $0$ handled separately in step 1.1.

If $n=0$, then ${\mathrm{\partial }}_0=0$ by [L2], so both composites from $C_0(X;G)$ to $0$ are the zero map.

Assume $n\ge 1$ and let $\sigma :{\mathrm{\Delta }}^n\to X$ be a singular $n$-simplex. By [L1] and [L2], ${\mathrm{\partial }}_n{f}_{\mathrm{\#},n}(\sigma )={\sum }_{i=0}^n(-1{)}^i(f\circ \sigma )\circ {\delta }_i={\sum }_{i=0}^n(-1{)}^{i}f\circ (\sigma \circ {\delta }_i)\mathrm{.}$ The same formulas give $f_{\mathrm{\#},n-1}{\mathrm{\partial }}_n(\sigma )=f_{\mathrm{\#},n-1}({\sum }_{i=0}^n(-1{)}^i\sigma \circ {\delta }_i)={\sum }_{i=0}^n(-1{)}^{i}f\circ (\sigma \circ {\delta }_i),$ so the two values agree on every singular simplex.

Singular simplices generate $C_n(X;\mathbb{Z})$, and the coefficient-$G$ map is the tensor extension of the integer-coefficient map. Therefore step 1.2 proves the identity for every $n\ge 1$, while step 1.1 handles degree $0$.

For every singular simplex $\sigma$ in $X$, [L1] gives ${\mathrm{id}}_{X,\mathrm{\#}}(\sigma )={\mathrm{id}}_X\circ \sigma =\sigma$ and $(g\circ f{)}_{\mathrm{\#}}(\sigma )=(g\circ f)\circ \sigma =g\circ (f\circ \sigma )=g_{\mathrm{\#}}(f_{\mathrm{\#}}(\sigma ))$. By linearity, ${\mathrm{id}}_{X,\mathrm{\#}}$ is the identity chain map and $(g\circ f{)}_{\mathrm{\#}}=g_{\mathrm{\#}}\circ f_{\mathrm{\#}}$ on singular chains.

By [L2], every induced map $f_{\mathrm{\#}}$ is a chain map. Therefore step 1.1 gives identity and composition laws in the category of chain complexes, and [L3] transfers those same laws to singular homology in each degree.

Let $\sigma :{\mathrm{\Delta }}^1\to X$ be a singular $1$-simplex and let $g\in G$. By [L2], ${\mathrm{\partial }}_1(\sigma \otimes g)=(\sigma {\delta }_0)\otimes g-(\sigma {\delta }_1)\otimes g\mathrm{.}$ Applying [L1] gives ${\epsilon }_X{\mathrm{\partial }}_1(\sigma \otimes g)=g-g=0.$

The generators $\sigma \otimes g$ span $C_1(X;G)$, so step 1.1 proves ${\epsilon }_X{\mathrm{\partial }}_1=0$ on all of $C_1(X;G)$. Therefore every $1$-boundary lies in $\ker {\epsilon }_X$, which is exactly the compatibility needed in degree $0$ for the reduced chain complex.

Let $\mathrm{\Phi }:C_0(X;\mathbb{Z})\to {⨁}_{C\in {\pi }_0^{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}(X)}\mathbb{Z}{e}_C$ send each singular $0$-simplex $\sigma$ to the basis vector indexed by the path component of its unique image point. If $\tau :{\mathrm{\Delta }}^1\to X$ is a singular $1$-simplex, then the map $s\mapsto \tau (1-s,s)$ is a path in $X$ from the image point of $\tau {\delta }_1$ to the image point of $\tau {\delta }_0$. Hence $\mathrm{\Phi }({\mathrm{\partial }}_1\tau )=0$, so $\mathrm{\Phi }$ kills $B_0(X;\mathbb{Z})$ and descends to a homomorphism $\bar{\mathrm{\Phi }}:H_0^{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}(X;\mathbb{Z})\to {⨁}_{C\in {\pi }_0^{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}(X)}\mathbb{Z}{e}_C\mathrm{.}$

The map $\bar{\mathrm{\Phi }}$ is surjective: if $X=\mathrm{\varnothing }$ there is nothing to prove, and otherwise any finite sum ${\sum }_{k=1}^m{a}_k{e}_{C_k}$ is the image of the $0$-cycle ${\sum }_{k=1}^m{a}_k{\sigma }_k$, where ${\sigma }_k$ is any singular $0$-simplex landing at a chosen point of $C_k$.

Let $z={\sum }_{k=1}^m{a}_k{\sigma }_k$ be a $0$-cycle with $\bar{\mathrm{\Phi }}([z])=0$. For each path component that meets the finite support of $z$, choose one base point simplex ${\sigma }_C$. Since $\bar{\mathrm{\Phi }}([z])=0$, the total coefficient in each such component is zero. If ${\sigma }_k$ and ${\sigma }_C$ lie in the same path component, [L2] gives a path ${\gamma }_k:I\to X$ from the image point of ${\sigma }_k$ to the image point of ${\sigma }_C$. Define the singular $1$-simplex ${\bar{\gamma }}_k:{\mathrm{\Delta }}^1\to X$ by ${\bar{\gamma }}_k(t_0,t_1)={\gamma }_k(t_1)$. Then ${\mathrm{\partial }}_1{\bar{\gamma }}_k={\sigma }_C-{\sigma }_k$. Therefore $[z]={\sum }_C\left({\sum }_{{\sigma }_k\in C}{a}_k\right)[{\sigma }_C]=0$ in $H_0^{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}(X;\mathbb{Z})$. Thus $\bar{\mathrm{\Phi }}$ is injective.

Steps 2.1 and 2.2 show that $\bar{\mathrm{\Phi }}$ is an isomorphism, so $H_0^{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}(X;\mathbb{Z})$ is free on the path components of $X$.

Since $X$ is nonempty and path-connected, it has exactly one path component. By [L1], $H_0^{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}(X;\mathbb{Z})\cong \mathbb{Z}\mathrm{.}$

Let $z={\sum }_{k=1}^m{a}_k{\sigma }_k\in C_0(X;\mathbb{Z})$. Under the isomorphism of step 1.1, the class $[z]$ maps to ${\sum }_{k=1}^m{a}_k\in \mathbb{Z}$, which is exactly ${\epsilon }_X(z)$. Therefore $z\in \ker {\epsilon }_X$ if and only if $[z]=0$ in $H_0^{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}(X;\mathbb{Z})$, equivalently if and only if $z\in B_0(X;\mathbb{Z})$. By [L3], every $0$-boundary lies in $\ker {\epsilon }_X$, so $\ker {\epsilon }_X=B_0(X;\mathbb{Z})\mathrm{.}$

By [L2], reduced degree-zero homology is ${\tilde{H}}_0^{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}(X;\mathbb{Z})=\ker {\epsilon }_X/\mathrm{im}{\mathrm{\partial }}_1=\ker {\epsilon }_X/B_0(X;\mathbb{Z})=0.$

The standard simplex ${\mathrm{\Delta }}^n$ is path-connected: if $u=(u_0,\dots ,u_n)$ and $v=(v_0,\dots ,v_n)$ are points of ${\mathrm{\Delta }}^n$, the straight-line map $t\mapsto (1-t)u+tv$ stays in ${\mathrm{\Delta }}^n$. Therefore any singular simplex $\sigma :{\mathrm{\Delta }}^n\to X$ has connected image. Since the sets $X_{\alpha }$ are pairwise disjoint and clopen by [L2], the image of $\sigma$ lies in exactly one summand $X_{\alpha }$.

Step 1.1 identifies $C_n(X;G)$ with the direct sum of the chain groups $C_n(X_{\alpha };G)$, degree by degree, and the singular boundary preserves the chosen summand because each face of a simplex in $X_{\alpha }$ still lands in $X_{\alpha }$. Thus $C_{\bullet }(X;G)\cong {⨁}_{\alpha \in A}{C}_{\bullet }(X_{\alpha };G)$ as chain complexes.

Cycles and boundaries of a direct-sum chain complex are taken componentwise, so homology also splits componentwise. Applying [L1] to the chain-complex isomorphism of step 2.1 yields $H_n^{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}(X;G)\cong {⨁}_{\alpha \in A}{H}_n^{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}(X_{\alpha };G)\mathrm{.}$

If $n=0$, then ${\mathrm{\Pi }}_0={\lambda }_0$ is the oriented edge from $(v_0,0)$ to $(v_0,1)$, so [L2] gives $\mathrm{\partial }{\mathrm{\Pi }}_0={\iota }_1-{\iota }_0$. This is exactly the displayed formula with no side-prism sum.

Assume $n\ge 1$. By [L2], each $\mathrm{\partial }{\lambda }_i$ is the alternating sum of its $n+2$ codimension-one faces. The face opposite $(v_0,0)$ in ${\lambda }_0=[(v_0,0),(v_0,1),\dots ,(v_n,1)]$ is the top face ${\iota }_1$, and the face opposite $(v_n,1)$ in ${\lambda }_n=[(v_0,0),\dots ,(v_n,0),(v_n,1)]$ is the bottom face ${\iota }_0$. For each $1\le i\le n$, the face of ${\lambda }_i$ opposite $(v_i,0)$ is the same $n$-simplex as the face of ${\lambda }_{i-1}$ opposite $(v_{i-1},1)$, so these interior faces occur twice in ${\sum }_{i=0}^n(-1{)}^i\mathrm{\partial }{\lambda }_i$ with opposite total signs and cancel.

The remaining uncancelled faces are the top face ${\iota }_1$ from ${\lambda }_0$, the bottom face $-{\iota }_0$ from ${\lambda }_n$, and the side faces obtained by deleting one vertex $v_j$ of ${\mathrm{\Delta }}^n$. For each fixed $j$, those side faces are exactly the prism simplices in the standard triangulation of ${\delta }_j({\mathrm{\Delta }}^{n-1})\times I$, and comparing the induced vertex order with the definition of ${\mathrm{\Pi }}_{n-1}$ gives the total contribution $-(-1{)}^j({\delta }_j\times {\mathrm{id}}_I{)}_{\mathrm{\#}}{\mathrm{\Pi }}_{n-1}$. Summing over $j$ and combining with steps 1.1 and 1.2 yields the stated boundary formula.

If $n=0$ and $\sigma :{\mathrm{\Delta }}^0\to X$ is a singular $0$-simplex, then [L2] gives $\mathrm{\partial }{\mathrm{\Pi }}_0={\iota }_1-{\iota }_0$. Composing with $H\circ (\sigma \times {\mathrm{id}}_I)$ and using [L1] and [L3] yields $\mathrm{\partial }{P}_H(\sigma )=g_{\mathrm{\#}}(\sigma )-f_{\mathrm{\#}}(\sigma )\mathrm{.}$

Assume $n\ge 1$ and let $\sigma :{\mathrm{\Delta }}^n\to X$ be a singular $n$-simplex. Compose the boundary formula [L2] with the continuous map $H\circ (\sigma \times {\mathrm{id}}_I):{\mathrm{\Delta }}^n\times I\to Y\mathrm{.}$ By [L1] and [L3], the image of ${\mathrm{\Pi }}_n$ is $P_H(\sigma )$, the image of ${\iota }_1$ is $g_{\mathrm{\#}}(\sigma )$, and the image of ${\iota }_0$ is $f_{\mathrm{\#}}(\sigma )$.

Again using [L1], [L3], and [L4], the image of the side-prism term ${\sum }_{j=0}^n(-1{)}^j({\delta }_j\times {\mathrm{id}}_I{)}_{\mathrm{\#}}{\mathrm{\Pi }}_{n-1}$ under $H\circ (\sigma \times {\mathrm{id}}_I)$ is exactly $P_H(\mathrm{\partial }\sigma )$. Therefore step 1.2 becomes $\mathrm{\partial }{P}_H(\sigma )=g_{\mathrm{\#}}(\sigma )-f_{\mathrm{\#}}(\sigma )-P_H(\mathrm{\partial }\sigma ),$ or equivalently $g_{\mathrm{\#}}(\sigma )-f_{\mathrm{\#}}(\sigma )=\mathrm{\partial }{P}_H(\sigma )+P_H(\mathrm{\partial }\sigma )\mathrm{.}$

Singular simplices generate $C_n(X;\mathbb{Z})$, and the coefficient-$G$ version is obtained by tensor extension, so step 2.1 holds on all chains for every $n\ge 1$, while step 1.1 handles degree $0$. Hence the stated identities hold in all degrees.

Extend the prism operator by $P_{H,-1}=0$ on the zero group $C_{-1}(X;G)$. By [L1], $f_{\mathrm{\#}}-g_{\mathrm{\#}}=\mathrm{\partial }(-P_H)+(-P_H)\mathrm{\partial },$ so the family $-P_H$ satisfies the defining identity of [L2] for a chain homotopy from $f_{\mathrm{\#}}$ to $g_{\mathrm{\#}}$. Thus the two induced singular chain maps are chain-homotopic.

Applying [L3] to the singular chain complex yields $H_n(f_{\mathrm{\#}})=H_n(g_{\mathrm{\#}})$ for every $n\ge 0$.

By [L1], choose a homotopy inverse $g:Y\to X$ with $g\circ f\simeq {\mathrm{id}}_X$ and $f\circ g\simeq {\mathrm{id}}_Y$. Applying [L3] gives $H_n((g\circ f{)}_{\mathrm{\#}})={\mathrm{id}}_{H_n(X;G)},H_n((f\circ g{)}_{\mathrm{\#}})={\mathrm{id}}_{H_n(Y;G)}\mathrm{.}$

By [L2], $H_n((g\circ f{)}_{\mathrm{\#}})=H_n(g_{\mathrm{\#}})\circ H_n(f_{\mathrm{\#}}),H_n((f\circ g{)}_{\mathrm{\#}})=H_n(f_{\mathrm{\#}})\circ H_n(g_{\mathrm{\#}})\mathrm{.}$ Combining this with step 1.1 shows that $H_n(g_{\mathrm{\#}})$ and $H_n(f_{\mathrm{\#}})$ are two-sided inverses. Hence $H_n(f_{\mathrm{\#}})$ is an isomorphism.

By [L1], the identity map on $X$ is homotopic to the constant map at some point $x_0\in X$. Let $p:X\to \ast$ be the unique map and let $i:\ast \to X$ send the point of $\ast$ to $x_0$. Then $i\circ p$ is the constant map at $x_0$, so $i\circ p\simeq {\mathrm{id}}_X$, while $p\circ i={\mathrm{id}}_{\ast }$. Hence $X$ is homotopy equivalent to a point.

Apply [L3] to the map $p:X\to \ast$. It induces an isomorphism on every singular homology group, which is exactly the displayed conclusion.