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Let $A[n]$ be the nonnegative chain complex having $A$ in degree $n$, zero in every other degree, and zero differential. The simplicial Eilenberg–MacLane space isIts underlying simplicial set is a Kan complex, with ${\pi }_{n}K(A,n)\cong A$ and all other positive homotopy groups zero.

Regard the given short exact sequence as a degreewise short exact sequence of chain complexes concentrated in degree $n$. The inverse functor in the Dold–Kan correspondence is exact, so it produces a degreewise short exact sequence of simplicial abelian groupsThe last map is degreewise surjective and hence a Kan fibration. Its strict fiber is $K(A_1,n)$, and a strict fiber of a fibration computes the homotopy fiber. This proves the asserted homotopy fiber sequence of pointed Kan complexes.

Because every $C_i$ is free abelian, reduction modulo $n$ gives a short exact sequence of chain complexesIts long exact sequence in homology isIf $\bar{z}$ is a cycle modulo $n$, choose a lift $z\in C_i$. Then $dz=nw$ for some $w\in C_{i-1}$, and the connecting homomorphism is $\hat{\beta }[\bar{z}]=[w]$.

The coefficient sequencesimilarly givesAt chain level, if $z$ lifts a mod-$n$ cycle and $dz=nw$ modulo $n^2$, then the Bockstein homomorphism is $\beta [\bar{z}]=[\bar{w}]$.

There is a morphism from the first short exact sequence to the second whose three vertical maps are reduction modulo $n$, reduction modulo $n^2$, and the identity on $C\otimes \mathbb{Z}/n$. Naturality of connecting homomorphisms givesExactness of the first long exact sequence says $\hat{\beta }{\rho }_{\ast }=0$, and hence

For the standard cellular chain complex of Real projective space ${\mathbb{RP}}^3$, there is one copy of $\mathbb{Z}$ in degrees $0,1,2,3$, with $d_2=2$ and $d_1=d_3=0$. Modulo two all cellular differentials vanish, while the Bockstein is the identity from degree two to degree one and zero elsewhere. Therefore

Finally, Smith normal form decomposes a bounded chain complex of finitely generated free abelian groups, up to chain isomorphism and contractible summands, into one-term complexes $\mathbb{Z}[i]$ and two-term complexesin degrees $i+1$ and $i$. The former represents a free homology summand, and the latter represents $\mathbb{Z}/m$ in degree $i$. The two requested conclusions now follow from the next two parts.

The reversed dual module of a chain complex becomes a chain complex with the unsigned precomposition differential above. A functional on $C_{-i}$ is sent to a functional on $C_{1-i}$, lowering the reversed degree by one.

For finite free total chain complexes, evaluation identifies the underlying graded tensor product $X\otimes C^{\prime }$ with the graded Hom complex of chain complexes, where $X$ is the reversed dual chain complex. Under the precomposition-first Hom differential, conjugation by the sign $(-1{)}^{j(j-1)/2}$ makes this a chain isomorphism. The identity $\rho (j)-\rho (j-1)=j-1$ works for every integer degree. If the grading is unbounded and only degreewise finite, products on the Hom side need not equal direct sums on the tensor side.

Singular homology forms a chain complex from formal sums of continuous maps from standard simplices into a topological space. Its homology groups are invariant under homotopy.

The chain complex with $R$ in a single degree $n$ and zero differential. A map from it into a complex chooses a cycle in degree $n$.

The tensor product of homologically graded chain complexes has differential $D(x_p\otimes y)=d{x}_p\otimes y+(-1{)}^p{x}_p\otimes dy$. The sign makes the two mixed terms in $D^2$ cancel. Total degree is the sum of the two degrees.