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We consider the following exact sequences, first is a proyective resolution of $C$ and second is an extension $\xi$ of $A$ by $C$:

$P_2\xrightarrow {d_2}{P_1}\xrightarrow{d_1}P_0\rightarrow C\rightarrow 0$

$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$

We already know that there exist $\alpha : P_1\to A$ and $\beta : P_0\to B$, and the identity $1_C:C\to C$ making the resulting diagram commutative.

We recall $e(C,A)$ is the set of all equivalence classes of extensions of $A$ by $C$ and $\operatorname{Ext}^1(C,A) = \operatorname{Ker}(\operatorname{Hom}(d_2,A))/\operatorname{Im}(\operatorname{Hom}(d_1,A))$.

Then, we have the correspondence $\psi:e(C,A)\to \operatorname{Ext}^1(C,A),\psi [\xi ]= [\alpha ]$.

Now, how can I prove that if $\xi$ is a split extension then $\psi [\xi]=0$?

Thanks.

mdp
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user74411
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1 Answers1

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I assume you know that a split extension is isomorphic to $0\to A\to A\oplus C\to C\to 0$, where the maps are the canonical projection and inclusion.

I also assume you have already proved that $\psi$ is independent of the choice of $\alpha$ and $\beta$. Now in this case you can choose $\beta=\begin{pmatrix}0\\\varepsilon\end{pmatrix}$, where $\varepsilon$ is the map $P_0\to C$, and $\alpha=0$. Then of course $\psi[\xi]=[0]=0$.

Julian Kuelshammer
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