If there is an exact sequence of $R$-modules $0 \rightarrow A \stackrel{\alpha}{\longrightarrow} B \stackrel{\beta}{\longrightarrow} C \rightarrow 0$, then $\mathrm{pd}(B) \leq \mathrm{max}\{ \mathrm{pd}(A), \mathrm{pd}(C) \}$.
Here, $\mathrm{pd}(B)$ denotes the projective dimension of $R$-module $B$.
Let $(D,d)$ be a projective resolution of $A$, then there is an exact sequence:
$\cdots \stackrel{d_2}{\longrightarrow} D_1 \stackrel{d_1}{\longrightarrow} D_0 \stackrel{\epsilon}{\longrightarrow} A \longrightarrow 0.$
So, there is also a sequence,
$\cdots \stackrel{d_2}{\longrightarrow} D_1 \stackrel{d_1}{\longrightarrow} D_0 \stackrel{\alpha \epsilon}{\longrightarrow} B \longrightarrow 0.$
In this sequence, $\mathrm{Ker} \alpha \epsilon = \mathrm{Ker} \epsilon = \mathrm{Im}d_1$, because $\alpha$ is injective. But this sequence might not be exact, as $\alpha$ might not be surjective.
Let $(D',d')$ be a projective resolution of $C$, then there is an exact sequence:
$\cdots \stackrel{d_2'}{\longrightarrow} D_1' \stackrel{d_1'}{\longrightarrow} D_0' \stackrel{\epsilon'}{\longrightarrow} C \longrightarrow 0.$
$D_0'$ is projective, then there exists a module homomorphism $\phi: D_0' \rightarrow B$, such that $\epsilon' = \beta \phi$.
So,
$\cdots \stackrel{d_2'}{\longrightarrow} D_1' \stackrel{d_1'}{\longrightarrow} D_0' \stackrel{\phi}{\longrightarrow} B \longrightarrow 0.$
Here, $\mathrm{Ker} \phi$ might be contained properly in $\mathrm{Ker} \epsilon'$. Meanwhile, $\phi$ is not necessarily surjective.
Then, how can I piece together all the facts to get a projective resolution of $B$ in order to prove the conclusion?
Many thanks.