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EDIT: the following argument misinterprets question (2) as using internal homs of A-mod For any $A$-module $Q$, equip $A\oplus Q$ with the structure of an $A$-algebra such that $Q$ is an ideal of square zero. Then, $\mathrm{Hom}(M,Q)=\ker(\mathrm{Hom}(M,A\oplus Q)\to\mathrm{Hom}(M,A))$. This is functorial in $Q$, so you can apply Yoneda. |
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For any $A$-module $Q$, equip $A\oplus Q$ with the structure of an $A$-algebra such that $Q$ is an ideal of square zero. Then, $\mathrm{Hom}(M,Q)=\ker(\mathrm{Hom}(M,A\oplus Q)\to\mathrm{Hom}(M,A))$. This is functorial in $Q$, so you can apply Yoneda. |
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