# In the existence of a short exact sequence, the projective dimension of $B$ is less than the larger of projective dimensions of $A$ and $C$

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.

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Note that you haven't used the assumption on the dimensions of $A$ and $C$ which tells you that you can find projective resolutions $P_{\bullet}' \to A$ and $P_{\bullet}'' \to C$ of lengths $\leq l = \max{\{\operatorname{pd}{A},\operatorname{pd}{C}\}}$ of $A$, and $C$, respectively. The horseshoe lemma then tells you that there is a projective resolution $P_{\bullet} \to B$ with $P_{n} = P_{n}' \oplus P_{n}''$. As a consequence, $P_{n} = 0$ for $n \gt l$, hence $\operatorname{pd}{B} \leq l$.
The map $P_{0} \to B$ is obtained by lifting the map $P_{0}'' \to C$ over $B \to C$ using projectivity of $P_{0}''$ and composing $P_{0}' \to A \to B$. Use the five lemma to see that the map $P_{0} \to B$ thus obtained is actually surjective. Replace $A,B,C$ by the kernels of the maps $P_{0}' \to A$, $P_0 \to B$ and $P_0'' \to C$, note that these sit in a short exact sequence by the nine lemma and proceed by induction.