Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 20G (Algebraic Topology)

A chain homotopy from $g$ to $f$ consists of homomorphisms $s_n:C_n\to C_{n+1}^{\prime }$ such thatfor every $n$. If $z\in C_n$ is a cycle, $d_{n}z=0$, so $f_n(z)-g_n(z)=d_{n+1}^{\prime }{s}_n(z)$ is a boundary. Hence $f_n(z)$ and $g_n(z)$ represent the same homology class. Every homology class has a cycle representative, proving $\boxed{f_{\ast }=g_{\ast }}$ in every degree.

For a finite chain complex of finite-dimensional vector spaces, its Euler characteristic is $\chi (C)={\sum }_n(-1{)}^n\dim C_n$. For finitely generated free abelian groups, replace dimension by rank. Write $Z_n=\ker d_n$ and $B_n=\mathrm{im}{d}_{n+1}$. The identities $\mathrm{rank}{C}_n=\mathrm{rank}{Z}_n+\mathrm{rank}{B}_{n-1}$ and $\mathrm{rank}{Z}_n=\mathrm{rank}{H}_n+\mathrm{rank}{B}_n$ show by cancellation that $\chi (C)={\sum }_n(-1{)}^n\mathrm{rank}{H}_n$. An exact sequence thus has alternating rank sum zero.

Starting at the left, the three displayed sequences have alternating rank sums respectively $0,1,1$. Therefore, given that one is exact, the first sequence is the exact one. For completeness its necessary image ranks, successively, are $11,13,7,6,14,11$, all nonnegative and consistent with the adjacent module ranks. Split each middle free group as the incoming image plus a complementary outgoing image, and use identity maps on these summands, to realize an exact sequence of the first rank pattern. Vanishing of the alternating sum alone is not a general test for the exactness of unspecified maps; the supplied existence assertion is what selects the sequence.