这是一份umass麻省大学 MATH 411作业代写的成功案例
Proof. If every $A_{n}$ is injective, or if every $C_{n}$ is projective, then the sequence $0 \longrightarrow A_{n} \longrightarrow B_{n} \longrightarrow C_{n} \longrightarrow 0$ splits; hence the sequence
$$
0 \longrightarrow \operatorname{Hom}{R}\left(C{n}, G\right) \longrightarrow \operatorname{Hom}{R}\left(B{n}, G\right) \longrightarrow \operatorname{Hom}{R}\left(A{n}, G\right) \longrightarrow 0
$$
splits, in particular is short exact; then
$$
0 \longrightarrow \operatorname{Hom}{R}(\mathcal{C}, G) \longrightarrow \operatorname{Hom}{R}(\mathcal{B}, G) \longrightarrow \operatorname{Hom}_{R}(\mathcal{A}, G) \longrightarrow 0
$$
is a short exact sequence of complexes (of abelian groups), which is natural in $\varepsilon$ and $G$, and the result follows.
MMATH 411 COURSE NOTES ：
applied to $\alpha, \beta, \gamma$, yields an exact sequence
$$
\operatorname{Hom}{R}(A, J) \longrightarrow \operatorname{Hom}{R}(A, L) \longrightarrow \mathrm{LExt}{R}^{1}(A, B) \longrightarrow 0 . $$ So does the first row of the diagram: $$ \operatorname{Hom}{R}(A, J) \longrightarrow \operatorname{Hom}{R}(A, L) \longrightarrow \operatorname{RExt}{R}^{1}(A, B) \longrightarrow 0 .
$$