$\newcommand{\F}[1]{\mathcal{#1}}\DeclareMathOperator{\im}{im}$The following is an exercise from the book mentioned in the title:
If $0 \to M \to P \to A \to 0$ is exact with $P$ projective (or $\F{F}$-acyclic), show that $\F{L}_{i}\F{F}(A) \cong \F{L}_{i – 1}\F{F}(M)$ for $i \ge 2$, and that $\F{L}_{1}\F{F}(A)$ is the kernel of $\F{F}(M) \to \F{F}(P)$. More generally, show that if
\begin{equation*}
0 \to M_{m} \to P_{m} \to P_{m – 1} \to \cdots \to P_{0} \to A \to 0
\end{equation*}
is exact with $P_{i}$ projective (or $\F{F}$-acyclic), then $\F{L}_{i}\F{F}(A) \cong \F{L}_{i – m – 1}\F{F}(M_{m})$ for $i \ge m + 2$, and $\F{L}_{m + 1}\F{F}(A)$ is the kernel of $\F{F}(M_{m}) \to \F{F}(P_{m})$.Conclude that if $P \to A$ is an $\F{F}$-acyclic resolution of $A$, then $\F{L}_{i}\F{F}(A) = H_{i}(\F{F}(P))$.
(Notations: $\F{F}$ is a right exact functor and $\F{L}_{\ast}\F{F}$ are the left derived functors.)
I was able to do the first two parts under the assumption of $P_{\ast}$ being $\F{F}$-acyclic but I am having trouble showing the last part "Conclude that…".
My thoughts
If I assume $P \to A$ to be a projective resolution, then for any $m$, I can form the exact sequence
$$0 \to \ker(d_{m}) \to P_{m} \to P_{m – 1} \to \cdots \to P_{0} \to A \to 0.$$
Moreover, I have a projective resolution of $\ker(d_{m})$ as
$$\cdots \to P_{m + 2} \to P_{m + 1} \to \ker(d_{m}) \to 0.$$
Using the definition of $\F{L}_{\ast}$, I can now compute $\F{L}_{\ast}\F{F}(M)$ in terms of $H_{\ast}(\F{F}(P)$). I haven't exactly worked out the indices to see if it matches, but my issue is that without assuming projectivity I don't see any way of solving it.
Sidenote
(Can be ignored by those wanting to answer the question.) In case any future readers are stuck in the first two parts, here's an outline:
First assume that we have
\begin{equation*}
0 \to M \to P \to A \to 0.
\end{equation*}
For $i \ge 2$, we have the following segment from the associated long exact sequence
\begin{equation*}
\F{L}_{i}\F{F}(P) \to \F{L}_{i}\F{F}(A) \to \F{L}_{i – 1}\F{F}(M) \to \F{L}_{i – 1}\F{F}(P).
\end{equation*}
Since $i – 1 \ge 1$ and $P$ is $\F{F}$-acyclic, we see that the two ends above are $0$. Thus, $\F{L}_{i}\F{F}(A) \cong \F{L}_{i – 1}\F{F}(M)$, as desired. Similarly work out $\F{L}_{1}\F{F}(A)$.
Now, assume that we have a general exact sequence as
\begin{equation*}
0 \to M_{m} \to P_{m} \xrightarrow{\varphi_{m}} P_{m – 1} \xrightarrow{\varphi_{m – 1}} \cdots \xrightarrow{\varphi_{1}} P_{0} \xrightarrow{\varphi_{0}} A \to 0.
\end{equation*}
By breaking the sequence into short exact sequences as
\begin{align*}
0 \to \ker(\varphi_{m}) & \to P_{m} \to \im(\varphi_{m}) \to 0, \\
0 \to \ker(\varphi_{m – 1}) & \to P_{m – 1} \to \im(\varphi_{m – 1}) \to 0, \\
& \vdots \\
0 \to \ker(\varphi_{1}) & \to P_{1} \to \im(\varphi_{1}) \to 0, \\
0 \to \ker(\varphi_{0}) & \to P_{0} \to A \to 0,
\end{align*}
we now inductively apply the previous part by noting that $\im(\varphi_{k}) = \ker(\varphi_{k – 1})$ and $\ker(\varphi_{m}) = M_{m}$.
Best Answer
Ok sorry I misunderstood the question. Let $i\geq 0$ and put $P_{-1}=A$ and $P_{-2}=0$ so that what I write makes sense if $i=1$ or $i=0$.
Consider the resolution $$...\to P_{i+1}\to P_i\to P_{i-1}\to P_{i-2}\to...\to P_0\to A\to 0$$ and truncate : $$...\to P_{i+1}\to P_i\to M_{i-1}\to 0$$ $$0\to M_{i-1}\to P_{i-1}\to P_{i-2}\to...\to P_0\to A\to 0$$ so you have $M_{i-1}=\ker(P_{i-1}\to P_{i-2})=\operatorname{coker}(P_{i+1}\to P_i)$.
Now, you have proven that $L_iF(A)=\ker(L_0F(M_{i-1})\to L_0F(P_{i-1}))=\ker(F(M_{i-1})\to F(P_{i-1}))$.
But since $F$ is right exact, you have $F(M_{i-1})=\operatorname{coker}(F(P_{i+1})\to F(P_i))=F(P_i)/\operatorname{im}(F(P_{i+1}\to P_i))$.
Thus $L_iF(A)=\ker(F(M_{i-1})\to F(P_{i-1}))=\ker(F(P_i)/\operatorname{im}(F(P_{i+1}\to P_i))\to F(P_{i-1}))=H_i(F(P_\bullet))$.