Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 17 / 2

The chain map identity $d_{D}f=f{d}_C$ gives, for $(x,y)\in M_i$,Thus $M$ is a chain complex; it is a sign convention for the algebraic mapping cone.

Put $E_i=C_{i-1}$ with differential $d_E=d_C$. Inclusion into the second summand and projection onto the first give a short exact sequence of chain complexesIts connecting homomorphism is particularly explicit. If $x\in C_{i-1}$ is a chain cycle, lift it to $(x,0)\in M_i$ and take its boundary. The second component is $(-1{)}^{i}f(x)$, so the long exact sequence in homology containsIf every $H_i(M)$ vanishes, the exact sequence makes every $f_{\ast }$ both injective and surjective. Conversely, if every $f_{\ast }$ is an isomorphism, the map $H_i(D)\to H_i(M)$ has zero image, while $H_i(M)\to H_{i-1}(C)$ has zero image and zero kernel; hence $H_i(M)=0$.

One can also see the converse directly on chain cycles. If $(x,y)\in M_i$ is a chain cycle, then $d_{C}x=0$ and $d_{D}y=-(-1{)}^{i}f(x)$. Injectivity of the induced map on homology $f_{\ast }$ gives $x=d_{C}z$. Subtracting the chain boundary $d_f(z,0)$ leaves $(0,t)$ with $d_{D}t=0$. Surjectivity of the induced map on homology $f_{\ast }$ lets us choose a chain cycle $w\in C_i$ and $v\in D_{i+1}$ such thatThen $(0,t)=d_f(w,v)$. Thus every chain cycle in $M$ is a chain boundary, proving the mapping cone acyclicity criterion:

For each prime number $p$, reduction of the mapping cone is the mapping cone of the reduced chain map. Part (a), whose proof works over any coefficient ring, therefore givesEvery $M_i$ is a finitely generated free abelian group. Multiplication by $p$ consequently gives a short exact sequence of chain complexesThe neighboring terms $H_{i+1}(M\otimes {\mathbb{F}}_p)$ and $H_i(M\otimes {\mathbb{F}}_p)$ in its long exact sequence in homology both vanish. Thus multiplication by $p$ on $H_i(M)$ is an isomorphism for every prime number $p$.

The group $H_i(M)$ is a finitely generated abelian group, since it is a quotient of a subgroup of $M_i$. Write it, using the Fundamental theorem of finitely generated abelian groups, aswith $T_i$ finite. If $b_i>0$, multiplication by any prime number is not surjective. If $T_i\ne 0$, choose a prime number dividing its order; multiplication by that prime is not injective. Both contradict the preceding isomorphism. Hence $H_i(M)=0$ for every $i$.

Applying part (a) integrally now proves the stronger natural conclusionThis is prime coefficient detection of quasi-isomorphisms. The finite-generation hypothesis is used precisely when excluding nonzero groups on which every prime acts invertibly.