[proofplan] The proofs of homotopy invariance, the long exact sequence, Mayer--Vietoris, excision, and the cellular-to-singular isomorphism for $\mathbb{Z}$-coefficients all proceed through universal algebraic constructions on chain complexes. We show that replacing $C_\bullet(X; \mathbb{Z})$ by $C_\bullet(X; A) = C_\bullet(X; \mathbb{Z}) \otimes_{\mathbb{Z}} A$ carries each proof through without modification, because the tensor product $- \otimes_{\mathbb{Z}} A$ is a right-exact functor that preserves chain maps, chain homotopies, direct sums, and short exact sequences of free modules. The key observation is that singular chain groups $C_n(X; \mathbb{Z})$ are free abelian, so tensoring with $A$ preserves all exact sequences involving them (not just right-exactness, but full exactness). [/proofplan] [step:Define the chain complex with coefficients and verify it is a chain complex] For an abelian group $A$, define \begin{align*} C_n(X; A) = C_n(X; \mathbb{Z}) \otimes_{\mathbb{Z}} A, \end{align*} with boundary map $d_n^A = d_n \otimes \operatorname{id}_A: C_n(X; A) \to C_{n-1}(X; A)$. This is a chain complex: $d_{n-1}^A \circ d_n^A = (d_{n-1} \circ d_n) \otimes \operatorname{id}_A = 0 \otimes \operatorname{id}_A = 0$, since $d_{n-1} \circ d_n = 0$ in the integral chain complex. The homology groups are $H_n(X; A) = \ker(d_n^A) / \operatorname{im}(d_{n+1}^A)$. Since $C_n(X; \mathbb{Z})$ is a free abelian group (freely generated by the set of singular $n$-simplices $\sigma: \Delta^n \to X$), the tensor product $C_n(X; A) \cong \bigoplus_\sigma A$ is a direct sum of copies of $A$, one for each singular $n$-simplex. A typical element is a finite sum $\sum_\sigma \sigma \otimes a_\sigma$ with $a_\sigma \in A$. [/step] [step:Show that the tensor functor preserves chain maps and chain homotopies] **Chain maps:** If $f_\bullet: C_\bullet(X; \mathbb{Z}) \to C_\bullet(Y; \mathbb{Z})$ is a chain map ($d \circ f = f \circ d$), then $f_\bullet \otimes \operatorname{id}_A: C_\bullet(X; A) \to C_\bullet(Y; A)$ is a chain map: \begin{align*} (d \otimes \operatorname{id}_A) \circ (f \otimes \operatorname{id}_A) = (d \circ f) \otimes \operatorname{id}_A = (f \circ d) \otimes \operatorname{id}_A = (f \otimes \operatorname{id}_A) \circ (d \otimes \operatorname{id}_A). \end{align*} **Chain homotopies:** If $F_n: C_n(X; \mathbb{Z}) \to C_{n+1}(Y; \mathbb{Z})$ is a chain homotopy with $d \circ F + F \circ d = g - f$, then $F_n \otimes \operatorname{id}_A$ is a chain homotopy between $f \otimes \operatorname{id}_A$ and $g \otimes \operatorname{id}_A$: \begin{align*} (d \otimes \operatorname{id}_A) \circ (F \otimes \operatorname{id}_A) + (F \otimes \operatorname{id}_A) \circ (d \otimes \operatorname{id}_A) = (d \circ F + F \circ d) \otimes \operatorname{id}_A = (g - f) \otimes \operatorname{id}_A. \end{align*} Therefore: if two chain maps are chain homotopic over $\mathbb{Z}$, they remain chain homotopic over $A$. [/step] [step:Show that tensoring preserves exact sequences of free modules] In general, $- \otimes_{\mathbb{Z}} A$ is only right-exact. However, every submodule of a free abelian group is free. Since $C_n(X; \mathbb{Z})$ is free for all $n$, the subgroups $Z_n = \ker(d_n)$ and $B_n = \operatorname{im}(d_{n+1})$ are also free abelian. Every short exact sequence of the form \begin{align*} 0 \to K \to F \to Q \to 0 \end{align*} with $K$ free abelian splits (since free abelian groups are projective). A split exact sequence remains exact after tensoring with any abelian group $A$. In particular, the short exact sequence \begin{align*} 0 \to C_\bullet(A; \mathbb{Z}) \xrightarrow{i} C_\bullet(X; \mathbb{Z}) \xrightarrow{q} C_\bullet(X, A; \mathbb{Z}) \to 0 \end{align*} has all terms free abelian (the quotient $C_n(X)/C_n(A)$ is free since $C_n(A)$ is a direct summand of $C_n(X)$ — it is generated by a subset of the free basis). Tensoring with $A$ preserves exactness: \begin{align*} 0 \to C_\bullet(A; A) \xrightarrow{i \otimes \operatorname{id}} C_\bullet(X; A) \xrightarrow{q \otimes \operatorname{id}} C_\bullet(X, A; A) \to 0. \end{align*} [guided] Why does freeness matter so much? The functor $- \otimes_{\mathbb{Z}} A$ is right-exact (it always preserves surjections and cokernels) but can fail to preserve injections. For example, tensoring the injection $\mathbb{Z} \xrightarrow{\times 2} \mathbb{Z}$ with $\mathbb{Z}/2\mathbb{Z}$ gives the zero map $\mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$, which is not injective. However, if the kernel $K$ in a short exact sequence $0 \to K \to F \to Q \to 0$ is free, the sequence splits: the projection $F \to Q$ has a section $s: Q \to F$ because we can define $s$ on a basis of $Q$ and extend linearly. A split exact sequence $0 \to K \to K \oplus Q \to Q \to 0$ clearly remains exact after tensoring: $0 \to K \otimes A \to (K \oplus Q) \otimes A \to Q \otimes A \to 0$ is the same as $0 \to K \otimes A \to (K \otimes A) \oplus (Q \otimes A) \to Q \otimes A \to 0$. In singular homology, all chain groups are free (they are free abelian on the set of singular simplices), and all submodules of free abelian groups are free (this is a theorem of algebra). So every short exact sequence arising from inclusions of chain complexes splits, and tensoring with $A$ preserves it. [/guided] [/step] [step:Transfer each theorem to $A$-coefficients] We now verify that each major theorem carries over: **Homotopy invariance:** If $f \simeq g: X \to Y$, the proof constructs a chain homotopy $F_n: C_n(X) \to C_{n+1}(Y)$ between $f_\#$ and $g_\#$ using the prism operator. By the chain homotopy preservation from the second step, $F_n \otimes \operatorname{id}_A$ is a chain homotopy between $f_\# \otimes \operatorname{id}_A$ and $g_\# \otimes \operatorname{id}_A$. By [Chain Homotopic Maps Induce Equal Maps on Homology](/theorems/1923) (which is a purely algebraic statement about chain complexes, valid over any coefficients), $f_* = g_*: H_n(X; A) \to H_n(Y; A)$. **Long exact sequence of a pair:** The proof applies the [Snake Lemma](/theorems/1930) to the short exact sequence $0 \to C_\bullet(B) \to C_\bullet(X) \to C_\bullet(X, B) \to 0$. By the previous step, this remains exact after tensoring with $A$. The Snake Lemma is a purely algebraic result about chain complexes, so it applies to the tensored sequence and produces the long exact sequence for $H_n(X, B; A)$. **Mayer--Vietoris sequence:** The proof applies the Snake Lemma to $0 \to C_\bullet(A \cap B) \to C_\bullet(A) \oplus C_\bullet(B) \to C_\bullet^{\mathcal{U}}(X) \to 0$ and uses the [Small Simplices Theorem](/theorems/2255). The short exact sequence remains exact after tensoring (all terms are free). The Small Simplices Theorem provides a chain homotopy equivalence $C_\bullet^{\mathcal{U}}(X) \hookrightarrow C_\bullet(X)$; tensoring preserves this chain homotopy equivalence. The result follows. **Excision:** The proof shows that the inclusion $(X \setminus Z, B \setminus Z) \hookrightarrow (X, B)$ induces an isomorphism on homology by constructing chain homotopy inverses using barycentric subdivision. Tensoring with $A$ preserves these chain homotopy inverses. **Cellular-to-singular isomorphism:** The proof uses the long exact sequences of pairs $(X^n, X^{n-1})$ and the fact that the relative homology $H_i(X^n, X^{n-1})$ is concentrated in degree $n$. All of these long exact sequences and isomorphisms are preserved under tensoring. [guided] The unifying principle is that every proof in singular homology has two layers: 1. **Geometric/topological layer:** Constructing chain maps, chain homotopies, and short exact sequences of chain complexes. This uses the topology of $X$. 2. **Algebraic layer:** Deriving conclusions from the chain-level data (Snake Lemma, chain homotopy implies equal maps on homology, etc.). This is pure algebra. Tensoring with $A$ affects only the algebraic layer. Since all the chain groups are free abelian, the tensoring is exact, and all the algebraic machinery (Snake Lemma, chain homotopy arguments) works identically. The geometric constructions (prism operators, barycentric subdivision, excision) are performed at the $\mathbb{Z}$-level and then tensored. The same reasoning applies to cohomology $H^n(X; A) = H^n(\operatorname{Hom}(C_\bullet(X; \mathbb{Z}), A))$: the functor $\operatorname{Hom}(-, A)$ is left-exact, and applying it to split exact sequences of free modules preserves exactness. The proofs of homotopy invariance, long exact sequence, Mayer--Vietoris, and excision for cohomology all transfer to $A$-coefficients by the same argument. [/guided] [/step]