Chain Homotopy and the Homotopy Category - Examples

Define $s_0=1_{\mathbb{Z}}:C_0\to C_1$ and $s_n=0$ otherwise. In degree $1$, $d_2{s}_1+s_0{d}_1=0+1_{\mathbb{Z}}=1_{C_1},$ and in degree $0$, $d_1{s}_0+s_{-1}{d}_0=1_{\mathbb{Z}}+0=1_{C_0}\mathrm{.}$

Thus $1_C=ds+sd$, so [L1] makes $1_C$ homotopic to $0$. By [L2], the complex is contractible.

Let $f=1_C$ and $g=0$. By [L1], the chosen $s$ satisfies $f-g=1_C=ds+sd,$ so [L2] gives $f\simeq g$.

Nevertheless $f_0=1_{\mathbb{Z}}$ and $g_0=0$, so the degree-$0$ components are different. By [L3], these distinct maps still induce the same homology map.

A degree-$1$ map has only one possible nonzero component $u_0:\mathbb{Z}\to \mathbb{Z}$, so it is determined by an integer $m$. A degree-$0$ map has components $u_1,u_0$, hence is determined by $(a,b)\in {\mathbb{Z}}^2$, and a degree $-1$ map is determined by $u_1:\mathbb{Z}\to \mathbb{Z}$.

Substituting these components into [L1] gives ${\mathrm{\partial }}_1(m)=(m,m),{\mathrm{\partial }}_0(a,b)=b-a\mathrm{.}$ Therefore $\ker ({\mathrm{\partial }}_0)=\{(a,a)\}$, exactly the degree-$0$ chain maps from [L2].

The complex is exact because $\ker (v)=\{(a,0)\}$ equals $\mathrm{im}(u)$, and $u$ is injective while $v$ is surjective. The displayed formulas for $s_0$ and $s_1$ are compatible with the splitting $\mathbb{Z}\oplus \mathbb{Z}\cong Z_1\oplus Z_0\cong \mathbb{Z}\oplus \mathbb{Z}\mathrm{.}$

A direct calculation gives $u{s}_1+s_{0}v=1_{\mathbb{Z}\oplus \mathbb{Z}},s_{1}u=1_{\mathbb{Z}},v{s}_0=1_{\mathbb{Z}},$ so the identity map is null-homotopic. Hence [L1] applies and the complex is contractible.

The complex is acyclic: multiplication by $2$ is injective, reduction mod $2$ is surjective, and its kernel is $2\mathbb{Z}$, which is the image of the first map.

If the complex were contractible, its identity map would be null-homotopic. In degree $0$, that would force the surjection $\mathbb{Z}\to \mathbb{Z}\mathrm{/}2$ to have a section, so the short exact sequence would split. It does not split, so the complex is not contractible. Hence the displayed complex refutes the statement, exactly as [L2] warns; [L1] remains true as the forward implication.

Because the source and target are both acyclic, the zero map induces isomorphisms on all homology groups. Hence [L2] makes it a quasi-isomorphism.

If this map had a homotopy inverse, the source complex would be homotopy equivalent to the zero complex and therefore contractible, contradicting [L1]. So the map is not a chain homotopy equivalence, and the statement refuted is false. This is consistent with [L3], which gives only the forward implication.

Applying [L1] with $k=1$ gives the displayed terms of $C[1]$ and the shifted differentials $d_3^{C[1]}=-d_2^C=-2,d_2^{C[1]}=-d_1^C=0.$

The displayed homology identifications are the cases $n=3,2,1$ of [L2]. Thus the example shows every sign and every reindexing explicitly.

By [L1], a degree-$0$ element $(a,b)$ lies in $\ker ({\mathrm{\partial }}_0)$ exactly when $b-a=0$, so $\ker ({\mathrm{\partial }}_0)=\{(a,a)\}$. The same formula shows $(a,a)={\mathrm{\partial }}_1(a)$, hence $\ker ({\mathrm{\partial }}_0)=\mathrm{im}({\mathrm{\partial }}_1)\mathrm{.}$

Therefore $H_0(\underset{}{\mathrm{Hom}}(C,C))=0$. By [L2], applied in the abelian category $\mathbf{A}\mathbf{b}$, this means ${\mathrm{Hom}}_{K(\mathbf{A}\mathbf{b})}(C,C)=0,$ so every endomorphism class of $C$ is zero in the homotopy category.