= Solution
Repeated <Smith normal form> puts a chain complex of finitely generated free abelian groups into its <elementary decomposition of a finite free chain complex>: one-term summands $\mathbb Z$ and two-term summands $0\to\mathbb Z\xrightarrow{m}\mathbb Z\to0$. Applying $\operatorname{Hom}_{\mathbb Z}(-,\mathbb Z)$ reverses a two-term summand but keeps the same multiplication by $m$. Reading its homology gives the <universal coefficient theorem for cohomology>
$$
0\longrightarrow\operatorname{Ext}_{\mathbb Z}^1(H_{n-1}(C),\mathbb Z)
\longrightarrow H^n(C;\mathbb Z)
\longrightarrow\operatorname{Hom}_{\mathbb Z}(H_n(C),\mathbb Z)
\longrightarrow0,
$$
split noncanonically. Thus if $H_n(C)\cong\mathbb Z^{b_n}\oplus T_n$ with $T_n$ finite, then
$$
H^n(C;\mathbb Z)\cong\mathbb Z^{b_n}\oplus T_{n-1}.
$$
The <universal coefficient theorem for homology> with $\mathbb F_p$ coefficients gives
$$
0\to H_n(C;\mathbb Z)\otimes\mathbb F_p\to H_n(C;\mathbb F_p)\to\operatorname{Tor}_1^{\mathbb Z}(H_{n-1}(C;\mathbb Z),\mathbb F_p)\to0.
$$
If the middle group vanishes for every prime $p$, so does the tensor term. Any nonzero free summand survives for every $p$, and any nonzero finite summand survives for a prime dividing its order. Since integral homology is finitely generated, it must vanish in every degree. This is <detection of integral acyclicity modulo primes>.
For the displayed complex, write
$$
d_M(c,d)=(-d_Cc,-f_\#c+d_Dd).
$$
Using $f_\#d_C=d_Df_\#$, one obtains
$$
d_M^2(c,d)=\bigl(0,f_\#d_Cc-d_Df_\#c\bigr)=0.
$$
Thus $M$ is the <mapping cone> of $f_\#$ in this sign convention. Its <long exact sequence in homology> shows that $f_*$ is an isomorphism exactly when $H_*(M)=0$. If $f_*$ is an isomorphism with $\mathbb F_p$ coefficients for every prime, then $M\otimes\mathbb F_p$ is acyclic for every $p$. Prime-field detection makes $M$ integrally acyclic, so the integral long exact sequence gives
$$
\boxed{f_*:H_*(C;\mathbb Z)\xrightarrow{\sim}H_*(D;\mathbb Z).}
$$
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