Prove that for a SES \(0\into A\into B\into C\), the group \(\ext(C,A)\) classifies extensions of \(C\) by \(A\) up to isomorphism.
Solution
1. An extension of \(C\) by \(A\) is a short exact sequence\[ 0\to A\xrightarrow{i}B\xrightarrow{p}C\to0, \]and two extensions are equivalent when there is an isomorphism of their middle terms commuting with the maps from \(A\) and to \(C\).
This is the standard Yoneda notion of equivalence of extensions.
2. Choose a free resolution\[ 0\to K\xrightarrow{j}F\xrightarrow{q}C\to0. \]Pulling an extension back along \(q:F\to C\) produces a split extension of \(F\) by \(A\).
Because \(F\) is free, hence projective, the surjection from the pullback middle term onto \(F\) admits a section.
3. A choice of splitting determines a homomorphism \(\phi:K\to A\), well-defined modulo maps \(K\to A\) extending to \(F\).
Lift the identity of \(F\) to the pullback middle term. Restricting this lift to \(K=\ker q\) lands in the copy of \(A\), giving \(\phi\). Changing the splitting by a map \(F\to A\) changes \(\phi\) by its restriction to \(K\).
4. Therefore an extension determines a class\[ [\phi]\in \operatorname{coker}\bigl(\operatorname{Hom}(F,A)\to\operatorname{Hom}(K,A)\bigr) =\operatorname{Ext}^1(C,A). \]
Applying \(\operatorname{Hom}(-,A)\) to the length-one free resolution identifies this cokernel with \(\operatorname{Ext}^1(C,A)\).
5. Conversely, a homomorphism \(\phi:K\to A\) defines an extension by the pushout\[ B_\phi=(A\oplus F)/\langle(\phi(k),-j(k)):k\in K\rangle. \]Then\[ 0\to A\to B_\phi\to C\to0 \]is exact.
The map \(A\to B_\phi\) is induced by the first summand, and \(B_\phi\to C\) is induced by \(q:F\to C\). The defining relations identify exactly the kernel coming from \(K\), giving exactness.
6. Two maps \(\phi,\phi'\) give equivalent extensions exactly when their difference extends to a map \(F\to A\).
If \(\phi'-\phi=h\circ j\), then \((a,f)\mapsto(a+h(f),f)\) descends to an isomorphism \(B_\phi\to B_{\phi'}\) commuting with \(A\) and \(C\). Conversely, an equivalence of extensions yields such a change of splitting after pullback to \(F\).
7. Hence equivalence classes of extensions of \(C\) by \(A\) are naturally in bijection with\[ \boxed{\operatorname{Ext}^1(C,A).} \]Under this bijection the Baer sum of extensions corresponds to addition in \(\operatorname{Ext}^1(C,A)\).
Steps <1>2–<1>6 construct mutually inverse assignments. The compatibility with addition is the standard pushout/pullback description of the Baer sum.