Singular Chains and Singular Homology - Examples

Every map ${\mathrm{\Delta }}^n\to \ast$ is the same constant map $c_n$, so each chain group is free of rank one on $c_n$. By [L2], ${\mathrm{\partial }}_0=0$, and for $n\ge 1$ one has ${\mathrm{\partial }}_n(c_n)={\sum }_{i=0}^n(-1{)}^i{c}_{n-1},$ which is $0$ for odd $n$ and $c_{n-1}$ for even $n$.

Therefore $\ker {\mathrm{\partial }}_n=0$ for even $n\ge 2$ and $\ker {\mathrm{\partial }}_n=\mathbb{Z}[c_n]=\mathrm{im}{\mathrm{\partial }}_{n+1}$ for odd $n\ge 1$, so $H_n^{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}(\ast ;\mathbb{Z})=0$ for all $n>0$. Also $H_0^{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}(\ast ;\mathbb{Z})\cong \mathbb{Z}[c_0]\cong \mathbb{Z}$. Since the augmentation of [L1] sends $c_0$ to $1$, its kernel is $0$, so the reduced degree-zero group also vanishes.

By [L1], the two face maps of ${\mathrm{\Delta }}^1$ pick out the terminal and initial vertices of the interval. Therefore [L2] gives $\mathrm{\partial }{\iota }_1={\iota }_1{\delta }_0-{\iota }_1{\delta }_1=v_1-v_0\mathrm{.}$

Again by [L1], the three face maps of ${\mathrm{\Delta }}^2$ are the affine inclusions of the three edges opposite $v_0$, $v_1$, and $v_2$. Applying [L2] gives the displayed alternating sum $\mathrm{\partial }{\iota }_2={\iota }_2{\delta }_0-{\iota }_2{\delta }_1+{\iota }_2{\delta }_2\mathrm{.}$

Expanding [L1] once more gives ${\mathrm{\partial }}^2{\iota }_2=\mathrm{\partial }(v_1{v}_2)-\mathrm{\partial }(v_0{v}_2)+\mathrm{\partial }(v_0{v}_1)=(v_2-v_1)-(v_2-v_0)+(v_1-v_0),$ where each edge boundary is the degree-one formula from the previous example.

The three positive and three negative vertex terms cancel pairwise, so the sum is $0$, exactly as predicted by [L2].

By [L2], $P_H({\iota }_1)$ is the sum of the two oriented triangles obtained from the standard triangulation of the square $I\times I$. Applying [L3] to the singular simplex ${\iota }_1$ gives $\mathrm{\partial }{P}_H({\iota }_1)=\beta -\alpha -P_H(\mathrm{\partial }{\iota }_1)\mathrm{.}$

The chain $\mathrm{\partial }{\iota }_1$ is the terminal vertex minus the initial vertex. By [L1], these two vertices trace the constant singular $1$-simplices $c_1$ and $c_0$, respectively, so $P_H(\mathrm{\partial }{\iota }_1)=c_1-c_0\mathrm{.}$ Substitution into step 1.1 gives $\mathrm{\partial }{P}_H({\iota }_1)=\beta -\alpha -c_1+c_0\mathrm{.}$ Constant singular $1$-simplices are genuine generators here, so they cannot be discarded in the unnormalized complex.

The interval $I$ is a nonempty convex subset of $\mathbb{R}$, so [L1] makes it contractible.

By [L2], $I$ has the same singular homology groups as a point. Substituting the computation from [L3] gives the displayed formulas for ordinary and reduced singular homology.

By [L1], the sphere inclusion $S^{n-1}\hookrightarrow {\mathbb{R}}^n\setminus \{0\}$ is a deformation retract inclusion.

Applying [L2] to that inclusion gives isomorphisms on singular homology in every degree, which is exactly the displayed statement.

By [A1], the spaces $X$ and $Y$ have isomorphic singular homology groups in every degree, so the refuted claim's hypothesis holds for this pair.

Write $W_2:=S^1\vee S^1$ and let $q:W_2\bigsqcup S^2\to Y$ be the quotient map from [L7], with both basepoints identified to $y_0\in Y$. Choose a small open disk $D\subseteq S^2$ around the sphere basepoint and small open arcs $O_1,O_2$ around the wedge point in the two circle summands of $W_2$. Put $U:=q(W_2\bigsqcup D),V:=q((O_1\cup O_2)\bigsqcup S^2)\mathrm{.}$ By the quotient description in [L7], these are open path-connected subsets of $Y$ covering $Y$. The set $U$ deformation retracts onto $W_2$ by contracting $D$ to $y_0$, the set $V$ deformation retracts onto $S^2$ by contracting the two arcs to $y_0$, and the overlap $U\cap V=q((O_1\cup O_2)\bigsqcup D)$ deformation retracts onto $y_0$.

By [L1], ${\pi }_1(X,([0],[0]))\cong {\mathbb{Z}}^2$. The group ${\mathbb{Z}}^2$ is abelian, while $F_2$ is not abelian because the reduced words $ab$ and $ba$ are distinct. Hence ${\pi }_1(X,([0],[0]))\cong ̸{\pi }_1(Y,y_0)$.

By [L2] and [L5], the deformation retraction in step 1.2 gives ${\pi }_1(U,y_0)\cong F_2$. By [L3] and [L5], the corresponding deformation retraction gives ${\pi }_1(V,y_0)=1$. Since $U\cap V$ deformation retracts to the point $y_0$, it is simply connected. Applying [L4] to the open cover $Y=U\cup V$ therefore yields ${\pi }_1(Y,y_0)\cong {\pi }_1(U,y_0)\ast {\pi }_1(V,y_0)\cong F_2\mathrm{.}$

If $X$ and $Y$ were homotopy equivalent, choose homotopy inverse maps $h:X\to Y$ and $k:Y\to X$, together with homotopies $H:k\circ h\simeq {\mathrm{id}}_X$ and $K:h\circ k\simeq {\mathrm{id}}_Y$. Put $\beta (t):=H(([0],[0]),t)$ and $\gamma (t):=K(y_0,t)$. [L6, step 1.3, construct] For any path $\rho$ from $a$ to $b$, the assignment $[\alpha ]\mapsto [\bar{\rho }\ast \alpha \ast \rho ]$ transports loop classes from ${\pi }_1(X,a)$ to ${\pi }_1(X,b)$, and reversing $\rho$ gives the inverse transport because the inserted pairs $\rho \ast \bar{\rho }$ and $\bar{\rho }\ast \rho$ contract to constant loops. Applying this to $\beta$ and $\gamma$ yields isomorphisms ${\beta }_{\mathrm{\#}}:{\pi }_1(X,k(h(([0],[0]))))\stackrel{\sim }{\to }{\pi }_1(X,([0],[0])),{\gamma }_{\mathrm{\#}}:{\pi }_1(Y,h(k(y_0)))\stackrel{\sim }{\to }{\pi }_1(Y,y_0)\mathrm{.}$ The maps $h:(X,([0],[0]))\to (Y,h(([0],[0])))$ and $k:(Y,y_0)\to (X,k(y_0))$ are pointed, so [L6] gives induced homomorphisms on fundamental groups. Precomposing $H$ and $K$ with based loops and then transporting the moving basepoints by ${\beta }_{\mathrm{\#}}$ and ${\gamma }_{\mathrm{\#}}$ shows that these induced homomorphisms are inverse isomorphisms up to the two basepoint transports. Therefore ${\pi }_1(X,([0],[0]))\cong {\pi }_1(Y,y_0)$, contradicting step 1.3.

Therefore $X$ and $Y$ are not homotopy equivalent even though step 1.1 shows that their singular homology groups agree.

The assignment $\phi (\sigma )=1$ on each singular $0$-simplex extends uniquely to a homomorphism from the free abelian group $C_0(I;\mathbb{Z})$ to $\mathbb{Z}$, so it is a singular $0$-cochain in the usual dual-group sense.

By [L2], each point $x\in I$ determines a singular $0$-simplex $c_x$, and distinct points give distinct maps, so $I$ has infinitely many singular $0$-simplices. The cochain $\phi$ takes the value $1$ on every one of them, so its support is infinite.

Thus $\phi$ is a singular cochain whose support is not finite, refuting the claim.