Chain Complexes and Homology — Examples

Since $d_n=0$, [L3] identifies the cycle inclusion with $1_{C_n}:C_n\to C_n$, so $Z_n(C)\cong C_n$. Likewise the image of $d_{n+1}=0$ is the zero object, so the boundary object is $B_n(C)\cong 0$.

Substituting those identities into [L2] gives $H_n(C)=C_n\mathrm{/}0\cong C_n\mathrm{.}$

In degree $1$, the outgoing differential is $f$, so $Z_1=\ker (f)$ and $B_1=0$. In degree $0$, the incoming differential is $f$ and the outgoing one is $0$, so $Z_0=N$ and $B_0=\mathrm{im}(f)$. Every other term is zero.

Therefore [L2] gives $H_1=\ker (f),H_0=N\mathrm{/}\mathrm{im}(f)=\mathrm{coker}(f),$ and $H_n=0$ for $n\mathrm{\ne }0,1$.

Multiplication by a nonzero integer on $\mathbb{Z}$ has zero kernel, because $mx=0$ forces $x=0$. Its cokernel is the quotient group $\mathbb{Z}\mathrm{/}m\mathbb{Z}$.

Applying [L1] to $f=\times m$ yields $H_1=0,H_0\cong \mathbb{Z}\mathrm{/}m\mathbb{Z}\mathrm{.}$

The displayed sequence is exact in $\mathbf{A}\mathbf{b}$: multiplication by $2$ is injective, the quotient map onto $\mathbb{Z}\mathrm{/}2\mathbb{Z}$ is surjective, and its kernel is the even subgroup.

By [L2], placing this exact sequence in consecutive degrees gives a chain complex that is exact at every nonzero term. Hence its homology vanishes in every degree, so it is acyclic.

The sequence is split: $r(a,b)=a$ is a retraction of $i$, and $s(c)=(0,c)$ is a section of $p$. This is exactly the structure in [L1].

With $h_1=r$ and $h_0=s$, one computes $h_{1}i=1_{\mathbb{Z}},i{h}_1+h_{0}p=1_{\mathbb{Z}\oplus \mathbb{Z}},p{h}_0=1_{\mathbb{Z}}\mathrm{.}$ So the identity on this complex is written degreewise as $dh+hd$, which is the promised contraction formula.

The chain condition holds because $(\times 4)\circ 1_{\mathbb{Z}}=(\times 2)\circ (\times 2)$. Also $Z_1(C)=Z_1(D)=0$, $B_0(C)=2\mathbb{Z}$, and $B_0(D)=4\mathbb{Z}$. Hence $H_1(C)=H_1(D)=0$, $H_0(C)\cong \mathbb{Z}\mathrm{/}2\mathbb{Z}$, and $H_0(D)\cong \mathbb{Z}\mathrm{/}4\mathbb{Z}$.

By [L2], the induced homology map comes from the degree-$0$ map $f_0=\times 2$ on cycles. It sends the boundary subgroup $2\mathbb{Z}$ into $4\mathbb{Z}$, so on quotients it is $\bar{x}⟼\overline{2x}\mathrm{.}$ In degree $1$ the homology groups are zero, so the induced map there is zero.

The target complex is acyclic, because both the kernel and cokernel of $1_{\mathbb{Z}}$ are zero. The source zero complex is acyclic as well, so the inclusion induces isomorphisms on all homology groups.

The target complex is nonzero while the source is zero, so the inclusion is not invertible. Thus this is a concrete counterexample, as asserted by [L2].

The complex is acyclic, so every homology group is zero. The maps $f$ and $g$ are distinct because their degree-$1$ components are $1_{\mathbb{Z}}$ and $0$.

Since all homology groups vanish, both induced homology maps are zero in every degree. Therefore $f\mathrm{\ne }g$ but they induce the same homology map, as claimed in [L2].

The family $S_{\bullet }$ is a subcomplex: the only nontrivial check is that $d_1(S_1)=0$ lands in $2\mathbb{Z}=S_0$. Thus [L1] applies.

By [L2], the quotient has terms $C_1\mathrm{/}{S}_1\cong \mathbb{Z}$ and $C_0\mathrm{/}{S}_0\cong \mathbb{Z}\mathrm{/}2\mathbb{Z}$. The descended differential is zero because $2x$ lies in $2\mathbb{Z}$ for every representative $x\in \mathbb{Z}$, so changing representatives does not change the class. Hence the quotient complex is as displayed.

The complex is split exact, hence acyclic, so every homology group is zero. Its chain groups have ranks $1$, $2$, and $1$. Therefore $\chi (C)=1-2+1=0,$ while the alternating sum of homology ranks is also $0$.

This agrees with the theorem [L2], so the example is a direct Euler-Poincare computation in a finite free complex.