Long exact sequence in etale cohomology with support Let $\Lambda := \mathbb{Z}/m\mathbb{Z}$, $X$ a scheme, $i : Y\hookrightarrow X$ a closed immersion, $j : U := X-Y\hookrightarrow X$ the open immersion of the complement, and $\mathcal{F}$ an etale sheaf of $\Lambda$-modules on $X$.
In Freitag-Kiehl p106 (Section 10 of Chapter 1), he describes the functors
$$\Gamma_Y : (\Lambda\text{-mod})(X)\rightarrow\text{Ab}\qquad \Gamma_Y(\mathcal{F}) := Ker(\mathcal{F}(X)\rightarrow\mathcal{F}(U))$$
and
$$\textbf{L}_Y : (\Lambda\text{-mod})(X)\rightarrow(\Lambda\text{-mod})(X)\qquad \textbf{L}_Y(\mathcal{F}) := Ker(\mathcal{F}\rightarrow j_*j^*\mathcal{F})$$
They are both left exact, and he writes $H_Y^i(X,\mathcal{F}) = R^i\Gamma_Y(\mathcal{F})$.
On p107, as shown here, he says that "we have the usual long exact sequences":
$$0\rightarrow\textbf{L}_Y(\mathcal{F})\rightarrow\mathcal{F}\rightarrow j_*j^*\mathcal{F}\rightarrow R^1\textbf{L}_Y(\mathcal{F})\rightarrow 0\qquad\qquad(*)$$
and $R^ij_*j^*\mathcal{F}\cong R^{i+1}\textbf{L}_Y(\mathcal{F})$ for $i\ge 1$.
I'm pretty confused how he gets this. I can't see a short exact sequence for which $(*)$ is the first few terms of the associated long exact sequence.
 A: Okay, so as usual the stacks project comes to the rescue.
The usual way of producing long exact sequences (say, to the right) is by applying a left exact functor $F$ to a short exact sequence in some abelian category, from which we get the long exact sequence with the higher derived functors $R^iF$. However, another way of saying this is that a short exact sequence
$$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$$
in an abelian category $\mathcal{A}$ yields a distinguished triangle in the derived category $D(\mathcal{A})$, namely, this is the triangle represented by $A[0]\rightarrow B[0]\rightarrow C[0]\rightarrow A[1]$ where the last map is obtained using mapping cones, and where $A[0]$ denotes the complex supported in degree 0 with value $A$, and similarly for $B[0],C[0]$.
This, combined with the result that "taking 0-th cohomology" is a homological functor tells us that we have a long exact sequence in $\mathcal{A}$:
$$\cdots \rightarrow H^0(A[0])\rightarrow H^0(B[0])\rightarrow H^0(C[0])\rightarrow H^0(A[1])\rightarrow H^0(B[1])\rightarrow H^0(C[1])\rightarrow\cdots$$
In the usual setting, all this amounts to saying is that $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ is an exact sequence, which says nothing new. However, if $F:\mathcal{A}\rightarrow\mathcal{B}$ is a left exact functor between abelian categories where $\mathcal{A}$ has enough injectives, then the nontrivial statement is that since derived functors preserve distinguished triangles, $RF(A[0])\rightarrow RF(B[0])\rightarrow RF(C[0])\rightarrow RF(A[1])$ is also a distinguished triangle in $D(\mathcal{B})$, and hence we obtain a long exact sequence
$$\cdots\rightarrow H^0(RF(A[0]))\rightarrow H^0(RF(B[0]))\rightarrow H^0(RF(C[0]))\rightarrow H^0(RF(A[1]))\rightarrow\cdots$$
Unraveling definitions, replacing $A[0],B[0],C[0]$ with injective resolutions $I^\bullet, J^\bullet,K^\bullet$ supported in nonnegative degrees, this is precisely the long exact sequence
$$0\rightarrow H^0(I^\bullet)\rightarrow H^0(J^\bullet)\rightarrow H^0(K^\bullet)\rightarrow H^1(I^\bullet)\rightarrow H^1(J^\bullet)\rightarrow\cdots$$
of derived functors.
The description above shows that ultimately every long exact sequence comes from a distinguished triangle in the derived category. The usual long exact sequence is produced by applying a derived functor to a short exact sequence. However, in the problem at hand, a distinguished triangle will be produced in another way.
Namely, I claim that there is a distinguished triangle in the derived category given by:
$$R\textbf{L}_Y(\mathcal{F}[0])\rightarrow \mathcal{F}[0]\rightarrow R(j_*j^*)(\mathcal{F}[0])\rightarrow R\textbf{L}_Y(\mathcal{F}[0])[1]\qquad\qquad(**)$$
Indeed, we may define this triangle to be constructed as follows. First, let  $\mathcal{F}[0]\cong I^\bullet$ be an injective resolution, then the sequence
$$0\rightarrow\textbf{L}_Y(I^\bullet)\rightarrow I^\bullet\rightarrow j_*j^*I^\bullet\rightarrow 0$$
is exact (in the abelian category of complexes) since the second map is surjective (injective modules being flasque). Thus, this gives a triangle in the derived category. Since $I^\bullet$ is a complex of injectives, the triangle can be written as
$$R\textbf{L}_Y(I^\bullet)\rightarrow I^\bullet\rightarrow R(j_*j^*)(I^\bullet)\rightarrow R\textbf{L}_Y(I^\bullet)[1]\qquad (***)$$
and since derived functors respect quasi-isomorphisms, we define the triangle $(**)$ to be obtained from $(***)$ via the canonical isomorphisms $$R\textbf{L}_Y(I^\bullet)\cong R\textbf{L}_Y(\mathcal{F}[0]), \quad I^\bullet\cong \mathcal{F}[0], \quad Rj_*j^*I^\bullet\cong Rj_*j^*\mathcal{F}[0]$$
Thus, $(**)$ is a distinguished triangle, which gives precisely the long exact sequence "$(*)$" alluded to on p107, and the "dimension shifting isomorphism" follows immediately from the fact that $\mathcal{F}[0]$ is supported in degree 0.
