Let $R$ be any ring. Show directly, without using Theorem $22.4$, that the following are equivalent.
i. Submodules of projective $R$-modules are projective.
ii. Homomorphic images of injective $R$-modules are injective.
For $(\mathrm{i}) \Rightarrow(\mathrm{ii})$, let $Q$ be an injective module and $U$ a submodule. Suppose $W \subseteq V$ and $\sigma: W \rightarrow Q / U$ are given. Map a projective module $P$ onto $V$ and let $P^{\prime}$ be the complete inverse image of $W .$ By assumption, $P^{\prime}$ is also projective. Now $\sigma$ extends to a map $\tau: P^{t} \rightarrow W \rightarrow Q / U$ and, since $P^{\prime}$ is projective and $Q \rightarrow Q / U$ is onto, $\tau$ lifts to a map $\tau^{*}: P^{\prime} \rightarrow Q$
Let $A_{R}$ and $B_{R}$ be $R$-modules, let
$$
\cdots \rightarrow P_{2} \stackrel{\alpha_{2}}{\longrightarrow} P_{1} \stackrel{\alpha_{1}}{\longrightarrow} P_{0} \rightarrow A \rightarrow 0
$$
be a projective resolution of $A$ and let $$
0 \rightarrow B \rightarrow Q_{0} \stackrel{\beta_{1}}{\longrightarrow} Q_{1} \stackrel{\beta_{2}}{\longrightarrow} Q_{2} \rightarrow \cdots
$$
be an injective resolution of $B$. Of course, the latter is a long exact sequence with each $Q_{i}$ injective. If we apply $\operatorname{Hom}_{R}(-, B)$ to the first sequence and $\operatorname{Hom}_{R}(A,-)$ to the second and then delete the $\operatorname{Hom}_{R}(A, B)$ term, we obtain the complexes
$$
\begin{aligned}
&0 \stackrel{\hat{\alpha}_{0}}{\longrightarrow} \operatorname{Hom}\left(P_{0}, B\right) \stackrel{\hat{\alpha}_{1}}{\longrightarrow} \operatorname{Hom}\left(P_{1}, B\right) \stackrel{\hat{\alpha}_{2}}{\longrightarrow} \operatorname{Hom}\left(P_{2}, B\right) \rightarrow \cdots \\
&0 \stackrel{\hat{\beta}_{0}}{\longrightarrow} \operatorname{Hom}\left(A, Q_{0}\right) \stackrel{\hat{\beta}_{1}}{\longrightarrow} \operatorname{Hom}\left(A, Q_{1}\right) \stackrel{\hat{\beta}_{2}}{\longrightarrow} \operatorname{Hom}\left(A, Q_{2}\right) \rightarrow \cdots
\end{aligned}
$$
where, for convenience, we let $\hat{\alpha}_{0}$ and $\hat{\beta}_{0}$ denote the appropriate zero maps. The abelian groups $\operatorname{Ext}_{R}^{n}(A, B)$ are now defined by
$$
\operatorname{Ext}_{R}^{n}(A, B)=\operatorname{Ext}^{n}(A, B)=\operatorname{Ker}\left(\hat{\beta}_{n+1}\right) / \operatorname{Im}\left(\hat{\beta}_{n}\right)
$$
for $n=0,1,2, \ldots$ It can be shown that Ext $^{n}(A, B)$ depends only on $A$ and $B$ and not on the particular injective resolution chosen for $B$. Furthermore,
$$
\operatorname{Ext}_{R}^{n}(A, B)=\operatorname{Ext}^{n}(A, B)=\operatorname{Ker}\left(\hat{\alpha}_{n+1}\right) / \operatorname{Im}\left(\hat{\alpha}_{n}\right)
$$