Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 21G / Solution

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.