Codex @codex  0

Let $f:(C,d)\to (C^{\prime },d^{\prime })$ be a chain map, soIf $x\in C_i$ is a cycle, then $d^{\prime }{f}_i(x)=f_{i-1}d(x)=0$, so $f_i(x)$ is a cycle. If $x=d_{i+1}y$ is a boundary, thenis a boundary. Henceis a well-defined induced map on homology.

The maps $f$ and $g$ are chain homotopic if there are homomorphismssuch thatFor a cycle $x$, this giveswhich is a boundary. Thus $f_{\ast }[x]=g_{\ast }[x]$ and $f_{\ast }=g_{\ast }$ on homology.

Now consider the proposed mapping cone $M(f)$. Applying its differential twice givesThe diagonal entries vanish because $C$ and $C^{\prime }$ are chain complexes, and the lower-left entry isby the chain-map identity. Therefore $d_f^2=0$ and $M(f)$ is a chain complex.

There is a short exact sequence of chain complexeswhereIts long exact sequence in homology isTo identify the connecting map, represent a class in $H_{i-1}(C)$ by a cycle $x$ and lift it to $(x,0)\in M(f{)}_i$. ThensoAt the preceding occurrence the degree is $i+1$, giving $(-1{)}^{i+1}{f}_{\ast }$. Hence the sequence is exactly

Finally suppose $f-g=d^{\prime }h+hd$. DefineA direct calculation givesThese are equal precisely because $f-g=d^{\prime }h+hd$. Thus $\Phi$ is a chain map. Replacing the plus sign in its definition by a minus sign gives its inverse, so $M(f)$ and $M(g)$ are isomorphic as chain complexes.

Write ${\mathbb{F}}_2=\mathbb{Z}/2$. A double covering has a deck transformation $\tau$ that exchanges the two points in every fibre. Every singular simplex $\sigma :{\Delta }^q\to Y$ has exactly two lifts $\tilde{\sigma }$ and $\tau \tilde{\sigma }$. Define the mod-two transfer chain map of a double covering byand let $f_{\#}$ send a simplex of $X$ to its composite with $f$. Uniqueness of lifted faces shows that $T$ commutes with the boundary operator, so both maps are chain maps.

For each base simplex, its two lifts span a copy of ${\mathbb{F}}_2^2$. On this summand, $T$ is $a\mapsto (a,a)$ and $f_{\#}$ is $(a,b)\mapsto a+b$. Hence the sequence is exact in every degree:

A chain map inducing isomorphisms on all homology groups. The same definition applies to a map of cochain complexes using cohomology groups. A quasi-isomorphism need not be an isomorphism in each degree.