I'm trying to do the following exercise from Weibel's Introduction to homological algebra regarding 'dimension shifting'
Exercise 2.4.3 : If $0\to M\to P\to A \to 0$ is exact with $P$ projective (or $F$-acyclic), show that $L_iF(A)\simeq L_{i-1}F(M)$ for $i\geq2$ and that $L_1F(A)$ is the kernel of $F(M)\to F(P)$.
$F$ is understood to be a right exact functor.
Technically the exercise has a second part slightly generalising the result but I believe understanding they key part I'm missing for the first part is enough to do the rest.
To prove what is asked, one gets from the short exact sequence a long exact sequence
\begin{align*}
\cdots&&\to &&L_{2}F(M)\to &&L_{2}F(P)\to &&L_{2}F(A)&& \\
&&\to &&L_{1}F(M)\to &&L_{1}F(P)\to &&L_{1}F(A)&& \\
&&\to &&F(M)\to &&F(P)\to &&F(A)&&\to0.
\end{align*}
and finally $P$ being $F$-acyclic we get the desired result.
At first, the long exact sequence seemed alright but then I thought about it and realised I need some clarifications. Here's how I tried to do it in much detail :
Take a projective resolution of $M$ and $A$, we then have the following diagram :
$\require{AMScd}$
\begin{CD}
@. @. @. 0 \\
@. @. @.@VVV \\
\cdots @>>> P'_1@>>> P'_0@>>> M @>>> 0\\
@. @.@. @VVV \\
@. @. @.P\\
@. @.@. @VVV \\
\cdots @>>> P''_1@>>> P''_0@>>> A @>>> 0\\
@. @. @.@VVV \\
@. @. @. 0 \\
\end{CD}
Using the horseshoe lemma we get a projective resolution of $P$ fitting into the diagram :
$\require{AMScd}$
\begin{CD}
@. 0 @. 0 @.0 \\
@. @VVV @VVV @VVV \\
\cdots @>>> P'_1@>>> P'_0@>>> M @>>> 0\\
@. @VVV@VVV @VVV \\
\cdots@>>> P_1@>>> P_0 @>>>P@>>>0\\
@. @VVV@VVV @VVV \\
\cdots @>>> P''_1@>>> P''_0@>>> A @>>> 0\\
@. @VVV @VVV@VVV \\
@. 0@. 0@. 0 \\
\end{CD}
And the sequence of complexes $0\to P'_{\bullet}\to P_{\bullet} \to P''_{\bullet}\to 0$ is exact. As I understand (and this might be where things go wrong) to obtain the long exact sequence we need, we would use the theorem stating that given a short exact sequence of chain complexes $0\to A \to B \to C\to 0$ one gets a long exact sequence on homology $\cdots \to H_{n+1}(C)\to H_{n}(A)\to H_n(B)\to H_n(C) \to H_{n-1}(A)\to\cdots $. Furthermore since by definition $L_iF(X) = H_i(F(Q))$ where $Q$ is a projective resolution of $X$, the short exact sequence we want needs to be $0\to F(P'_\bullet)\to F(P_{\bullet}) \to F(P''_{\bullet}) \to 0$. What I fail to understand is why should this last sequence be exact ? The functor $F$ is right exact, hence $ F(P'_\bullet)\to F(P_{\bullet}) \to F(P''_{\bullet}) \to 0$ is exact but this is not enough to get a long exact sequence on homology ? This is what I need to understand, how do we get that long sequence with the derived functors ?
Best Answer
Since $P''_n$ is projective, each short exact sequence $0 \to P'_n \to P_n \to P_n'' \to 0$ is split. Any additive functor will preserve this splitting, so $0\to F(P'_n) \to F(P_n) \to F(P''_n) \to 0$ is not only exact but also split.