= Solution
First check the signs directly. For $(x,y)\in M_i$, applying the proposed differential twice gives
$$
d_f^2(x,y)=\left(d_C^2x,\;(-1)^{i-1}f(d_Cx)+(-1)^i d_Df(x)+d_D^2y\right)=(0,0),
$$
because $f$ is a <chain map>. Thus $M$ is a <chain complex>.
Let $C'$ be the shifted <chain complex> with $C'_i=C_{i-1}$ and differential $d_C$; here there is no minus sign in that differential. Inclusion in the second summand and projection onto the first give a <short exact sequence of chain complexes>
$$
0\longrightarrow D\longrightarrow M\longrightarrow C'\longrightarrow0.
$$
To compute the connecting map in its <long exact sequence in homology>, lift a cycle $x\in C_{i-1}$ to $(x,0)\in M_i$. Its boundary is $(0,(-1)^if(x))$. Hence the connecting map $H_i(C')\to H_{i-1}(D)$ is $(-1)^if_*$, and the relevant exact portion is
$$
H_i(C)\xrightarrow{(-1)^{i+1}f_*}H_i(D)\longrightarrow H_i(M)\longrightarrow H_{i-1}(C)\xrightarrow{(-1)^if_*}H_{i-1}(D).
$$
If every $f_*$ is an <isomorphism>, exactness makes $H_i(M)=0$. Conversely, if every $H_i(M)$ is zero, the adjacent exact portions make every $f_*$ both injective and surjective. Therefore
$$
\boxed{H_*(M)=0\quad\Longleftrightarrow\quad f\text{ is a quasi-isomorphism}.}
$$
This is the acyclicity criterion for a <mapping cone>, with the degree-dependent signs adjusted to the given convention.
For the assertion about spaces, use the <cellular approximation theorem> to replace $f$ by a <cellular map>, and take the induced map of the finite free <cellular chain complexes>. Tensoring these complexes and their cone with $\mathbb F_p$ gives the corresponding mod-$p$ complexes and cone. The same exact-sequence argument works over $\mathbb F_p$, so the assumed <homology> isomorphisms imply
$$
H_i(M\otimes\mathbb F_p)=0\qquad\text{for every }i\text{ and every prime }p.
$$
The <universal coefficient theorem for homology> gives
$$
0\longrightarrow H_i(M)\otimes\mathbb F_p\longrightarrow H_i(M\otimes\mathbb F_p)\longrightarrow\operatorname{Tor}_1^{\mathbb Z}(H_{i-1}(M),\mathbb F_p)\longrightarrow0.
$$
In particular $H_i(M)\otimes\mathbb F_p=0$ for every prime. Each $H_i(M)$ is a <finitely generated abelian group>. A nonzero free summand would survive tensoring with every $\mathbb F_p$, while a nonzero finite cyclic summand would survive for a prime dividing its order. Thus $H_i(M)=0$ in every degree, by <detection of integral acyclicity modulo primes>. Applying the cone criterion once more proves \b[the integral homology map is an isomorphism in every degree]. Finite generation is what makes detection by all prime fields sufficient.
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