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Applying the contravariant Hom functor ${\mathrm{Hom}}_R(-,W)$ to $0\to I\to X\to C\to 0$ givesThe connecting map sends $I\to W$ to its pushout of a module extension. Its vanishing means that map extends to $X\to W$. For a hereditary ring, ${\mathrm{Ext}}^2(C,W)=0$, making the indicated restriction on first extension groups surjective. The covariant variable has its corresponding long exact sequence.

Choose a projective presentation $0\to K\stackrel{j}{\to }P\stackrel{\pi }{\to }C\to 0$, with $P$ a projective module. Continuing it to a projective resolution showsIndeed a degree-one cocycle in the Hom functor applied to the projective resolution descends to $K$, while the degree-one coboundaries are exactly restrictions of maps from $P$. This establishes the description of the Ext functor without first assuming the extension classification.

For a module extension with inclusion $i:A\to B$ and quotient map $r:B\to C$, projectivity lifts $\pi$ to $s:P\to B$. Since $rsj=0$, there is a unique map $h:K\to A$ with $ih=sj$. A different lift changes $h$ by $\ell j$ for some $\ell :P\to A$. An equivalence of module extensions, which is the identity on both end modules, also preserves this class. We have therefore defined a map from extension classes to ${\mathrm{Ext}}_R^1(C,A)$.

Conversely, for $h:K\to A$, form the pushout of a module extensionThe map $A\to B_h$ sends $a$ to $[a,0]$, and $B_h\to C$ sends $[a,p]$ to $\pi (p)$. The first is injective because $j$ is injective. If $\pi (p)=0$, write $p=j(k)$; then $[a,p]=[a+h(k),0]$, proving exactness in the middle. The last map is surjective. Thus $B_h$ is a module extension. If $h^{\prime }=h+\ell j$, the map $[a,p]\mapsto [a-\ell (p),p]$ is an equivalence of module extensions from $B_h$ to $B_{h^{\prime }}$. Finally $[a,p]\mapsto i(a)+s(p)$ identifies the pushout of a module extension built from an existing extension with its middle module. These two constructions are inverse, proving the classification:The zero class corresponds to a split short exact sequence. Fixed end modules matter: equivalence does not permit arbitrary automorphisms of $A$ or $C$.

In the second calculation the acting group is the infinite cyclic group. Its modules are modules over the group ring $S=\mathbb{Z}[z,z^{-1}]$, not merely over the underlying ring $\mathbb{Z}$. The trivial representation has $z$ acting as the identity. Its projective resolution isMultiplication by $z-1$ is injective, and the augmentation quotient is $\mathbb{Z}$. Applying the Hom functor into the trivial module makes the differential zero, so ${\mathrm{Ext}}_S^1(\mathbb{Z},\mathbb{Z})\cong \mathbb{Z}$. Taking two copies gives $\boxed{\mathcal{E}({\mathbb{Z}}^2,\mathbb{Z})\cong {\mathbb{Z}}^2}$.

An explicit representative of this extension of trivial modules for an infinite cyclic group is the abelian group $B_{m,n}=\mathbb{Z}u\oplus \mathbb{Z}v\oplus \mathbb{Z}w$, with injection $1\mapsto u$, quotient $v\mapsto (1,0)$, $w\mapsto (0,1)$, and actionThis is an invertible action: the inverse subtracts the same multiples of $u$. Every underlying abelian-group extension splits because ${\mathbb{Z}}^2$ is free, so any group-module extension has this form after choosing lifts of the quotient basis. Replacing those lifts by multiples of $u$ does not alter $m,n$. An equivalence fixing the ends has exactly such changes of lifts, so two representatives are equivalent precisely when their ordered pairs agree. Only $(0,0)$ splits as a group-module extension; overlooking the group action would incorrectly give a single class.