Step 1: Decomposition of the chain complex.

Let $C_n = C_n(X)$ be the singular chain groups, $Z_n = \ker \partial_n$, and $B_n = \operatorname{im}\, \partial_{n+1}$. Since $C_n$ is a free abelian group (it has a basis of singular $n$-simplices), and $Z_n \subseteq C_n$ is a subgroup of a free abelian group, $Z_n$ is also free. Consider the short exact sequences:

$0 \to Z_n \xrightarrow{\iota} C_n \xrightarrow{\partial_n} B_{n-1} \to 0 \quad (*)$ $0 \to B_n \xrightarrow{j} Z_n \to H_n(X) \to 0 \quad (**)$

Since $B_{n-1}$ is a subgroup of the free group $C_{n-1}$, it is free, so $(*)$ splits.

Step 2: Apply $\operatorname{Hom}(-, G)$ to $(*)$.

Since $(*)$ splits, applying $\operatorname{Hom}(-, G)$ gives an exact sequence: $0 \to \operatorname{Hom}(B_{n-1}, G) \xrightarrow{\partial_n^*} \operatorname{Hom}(C_n, G) \xrightarrow{\iota^*} \operatorname{Hom}(Z_n, G) \to 0$

Now $C^n(X; G) = \operatorname{Hom}(C_n, G)$ and the coboundary $\delta^{n-1}$ factors through $B_{n-1}$. We identify:

• $Z^n(X; G) = \ker \delta^n = \ker(\partial_{n+1}^* \circ -) = \operatorname{Hom}(Z_n, G)$ (via $\iota^*$)
• $B^n(X; G) = \operatorname{im}\, \delta^{n-1} \cong \operatorname{Hom}(B_{n-1}, G) / \ker \partial_n^*$

More precisely, using the identification $Z^n \cong \operatorname{Hom}(Z_n, G)$ under $\iota^*$, the coboundaries correspond to those homomorphisms $Z_n \to G$ that extend to $C_n \to G$ via $\partial_n^*$, which by the splitting means $B^n \cong \operatorname{im}(j^* : \operatorname{Hom}(Z_n, G) \to \operatorname{Hom}(B_n, G))$ pulled back.

Step 3: Apply $\operatorname{Hom}(-, G)$ to $(**)$.

The sequence $(**)$ gives rise to the exact sequence: $0 \to \operatorname{Hom}(H_n, G) \to \operatorname{Hom}(Z_n, G) \xrightarrow{j^*} \operatorname{Hom}(B_n, G) \to \operatorname{Ext}^1(H_n, G) \to 0$

since $Z_n$ is free (so $\operatorname{Ext}^1(Z_n, G) = 0$).

Step 4: Assemble the result.

We have $H^n(X; G) = Z^n / B^n \cong \operatorname{Hom}(Z_n, G) / \operatorname{im}(j^*)_{\text{restricted}}$. From Step 3, we can identify the kernel and cokernel to obtain:

$0 \to \operatorname{coker}(j^*_{n-1}) \to H^n(X;G) \to \ker(j^*_n \text{ restricted}) \to 0$

which yields exactly:

$0 \to \operatorname{Ext}^1(H_{n-1}(X), G) \to H^n(X; G) \xrightarrow{h} \operatorname{Hom}(H_n(X), G) \to 0$

The map $h$ sends a cocycle $\varphi : C_n \to G$ to its restriction $\varphi|_{Z_n}$, which factors through $H_n = Z_n / B_n$ precisely when $\varphi$ annihilates $B_n$.

Step 5: Splitting.

Since $Z_n$ is free, the sequence $(**)$ admits a section $s : Z_n \to Z_n$ (identity) factoring through a map $H_n \to Z_n$. Choose a splitting $\sigma : H_n(X) \to Z_n$ of $(**)$. Then $\sigma^* : \operatorname{Hom}(Z_n, G) \to \operatorname{Hom}(H_n, G)$ provides a right inverse for $h$, splitting the UCT sequence.

The splitting depends on the choice of $\sigma$, hence is not natural. $\square$