The definition of the cohomology of an Abelian sheaf This is how the cohomology of an Abelian sheaf is defined in Kempf's Algebraic Varieties:
Recall that we have the exact sequences
$0 \to \mathcal{F}\to D(\mathcal{F})\to D(\mathcal{F})/\mathcal{F} \to 0$
We may iterate this construction as follows: Let $C^0(\mathcal{F})=\mathcal{F}$, $D^i(\mathcal{F})=D(C^i(\mathcal{F}))$ and 
$$0 \to C^{i}(\mathcal{F}) \to D^i(\mathcal{F}) \to C^{i+1}(\mathcal{F})\to0$$
be the canonical quotient sequence. Putting these short exact sequences together (how?) we get a resolution
$$\mathcal{F} \to D^0(\mathcal{F}) \to D^1(\mathcal{F})\to \cdots$$
of $\mathcal{F}$ by the complex $D^*(\mathcal{F})$. The sheaves are $D^{i}(\mathcal{F})$ are flabby construction (why?) and the construction is functorial in $\mathcal{F}$. We get a complex $\Gamma(X,D^*\mathcal{F})$ by taking global sections. The $i$-homology group of this complex is the $i$-th cohomology group $H^i(X,\mathcal{F})$. Clearly, $H^i(X,-)$ is an additive functor. We have a natural mapping $\Gamma(X,\mathcal{F}) \to H^0(X,\mathcal{F})$ which is easily seen to be an isomorphism.
My questions: 
1) How is $C^{i+1}(\mathcal{F})$ defined? Is it the same as $D^{i}(\mathcal{F})/C^i(\mathcal{F})$? Because I can't see it from the iterative definition given by Kempf and I'm only guessing.
2) A complex is like a chain of objects and arrows $d_i$ such that $d_{i+1} \circ d_i = 0$. Am I right?
3) Why are the sheaves flabby by construction? What does the construction is functorial in $\mathcal{F}$ mean in this context?
4)  Why is the last mapping an isomorphism?
Please forgive me if my questions are very basic. I have no background in homology/cohomology theory and I think Kempf has assumed that the reader already knows these things because it hasn't explained this stuff at all.
 A: Question 0: The map $$ \delta_i :  D^{i}(\mathcal F) \to D^{i+1}(\mathcal F)$$ in the resolution is defined as the composition of the surjective map $D^{i}(\mathcal F) \to C^{i+1}(\mathcal F)$ with the injective map $C^{i+1}(\mathcal F) \to D^{i+1}(\mathcal F)$.
It is a standard exercise in diagram-chasing to verify that the resolution $$ 0 \to \mathcal F \to D^0(\mathcal F) \to D^1(\mathcal F) \to \dots$$
constructed in this way is indeed exact.
Question 1: $C^{i+1}(\mathcal F)$ is defined to be $D^{i}(\mathcal F)/ C^i(\mathcal F)$.
This is actually implicit from the short exact sequence $$0 \to C^{i}(\mathcal{F}) \to D^i(\mathcal{F}) \to C^{i+1}(\mathcal{F})\to 0 ,$$
which tells you that $C^{i+1}(\mathcal {F})$ is the cokernel of the injective map $C^{i}(\mathcal{F}) \to D^i(\mathcal{F})$; the only way this is possible is if $C^{i+1}(\mathcal F)$ is isomorphic to the quotient $D^{i}(\mathcal F)/ C^i(\mathcal F)$.
Question 2: Yes, a complex is a chain of objects from an abelian category with arrows $d_i$ such that $d_{i+1} \circ d_i = 0$. Since $$ D^0(\mathcal F) \to D^1(\mathcal F) \to D^2(\mathcal F) \to \dots$$ is an exact sequence of sheaves, and since the global sections function $\Gamma(X,-)$ is left-exact, it is a standard exercise to verify that $$\Gamma(X, D^0(\mathcal F)) \to \Gamma(X, D^1(\mathcal F)) \to \Gamma(X, D^2(\mathcal F)) \to \dots$$ is a complex.
Question 3a: Kempf is only claiming that the $D^{i}(\mathcal F)$'s are flabby. This is obvious from the definition of $D^i(F)$ as the sheaf of discontinuous sections of $\mathcal F$, which says that for any open $U \subset X$, the group of sections $D^i(\mathcal F)(U)$ is the set of maps $$D^i(\mathcal F)(U) = \left\{ s : U \to \prod_{p \in U} \mathcal F_p {\rm \ such \ that \ } s(p) \in \mathcal F_p {\rm \ for \ all \ } p \in U \right\}.$$ To check that $D^i(\mathcal F)$ is flabby, you need to check that for any open $V \subset U$, the restriction map $D^i(\mathcal F)(U) \to D^i(\mathcal F)(V)$ is surjective. It's not too hard to see that this is true.
Question 3b: The construction being functorial means that if you have a morphism of sheaves of abelian groups $\phi : \mathcal F \to \mathcal G$, then you can find maps $$\phi_i : D^i(\mathcal F) \to D^i(\mathcal G)$$ such that $$\delta^{\mathcal G}_i \circ \phi_i = \phi_{i + 1} \circ \delta^{\mathcal F}_i,$$ where $\delta_i^{\mathcal F} : D^{i}(\mathcal F) \to D^{i+1}(\mathcal F)$ and $\delta_i^{\mathcal G} : D^{i}(\mathcal G) \to D^{i+1}(\mathcal G)$ are the maps in the resolutions of $\mathcal F$ and $\mathcal G$.
Question 4: The global sections functor $\Gamma(X, - )$ is left-exact. So if you apply $\Gamma(X, - )$ to the exact sequence $ 0 \to \mathcal F \to D^0(\mathcal F) \to D^1(\mathcal F) $, you see that
$$ 0 \to \Gamma(X, \mathcal F) \to \Gamma(X, D^0(\mathcal F)) \to \Gamma(X, D^1(\mathcal F))$$
is exact, hence
$$ \Gamma(X, \mathcal F) \cong {\rm ker}\left(\Gamma(X, D^0(\mathcal F)) \to \Gamma(X, D^1(\mathcal F))\right) := H^0(X, \mathcal F)$$

Extra details:
(a) To see that the resolution in Question 0 is exact, pick an arbitrary point $p \in X$, and consider the maps on the stalks. First observe that $${\rm im}(D^i(\mathcal F)_p \to D^{i+1}(\mathcal F)_p) = {\rm im}(C^{i+1}(\mathcal F)_p \to D^{i+1}(F)_p)$$ since $D^i(\mathcal F)_p \to C^{i+1}(\mathcal F)_p$ is surjective.
Also, $${\rm ker}(D^{i+1}(\mathcal F)_p \to D^{i+2}(\mathcal F)_p) = {\rm ker}(D^{i+1}(\mathcal F)_p \to C^{i+2}(F)_p)$$ since $C^{i+2}(\mathcal F)_p \to D^{i+2}(F)_p$ is injective.
It follows that $${\rm im}( D^i(\mathcal F)_p \to D^{i+1}(\mathcal F)_p) = {\rm ker}(D^{i+1}(\mathcal F)_p \to D^{i+2}(\mathcal F)_p)$$ since ${\rm im}(C^{i+1}(\mathcal F)_p \to D^{i+1}(\mathcal F)_p)= {\rm ker}(D^{i+1}(\mathcal F)_p \to C^{i+2}(\mathcal F)_p)$. Since the point $p \in X$ was arbitrary, this shows that the resolution is exact.
(b) To see that the you get a complex in Question 2, use left-exactness of $\Gamma(X, - )$ to obtain exact sequences
$$ 0 \to \Gamma(X, C^{i}(\mathcal F)) \to \Gamma(X, D^{i}(\mathcal F)) \to \Gamma(X, C^{i+1}(\mathcal F))$$
[Note that the final map $\Gamma(X, D^{i}(\mathcal F)) \to \Gamma(X, C^{i+1}(\mathcal F))$ is not surjective!]
Then apply similar reasoning as in question 0 to see that $ {\rm im}(\Gamma(X, D^{i}(\mathcal F)) \to \Gamma(X, D^{i+1}(\mathcal F)))$ is a subgroup of ${\rm ker}(\Gamma(X, D^{i+1}(\mathcal F)) \to \Gamma(X, D^{i+2}(\mathcal F)))$.
[Here, $ {\rm im}(\Gamma(X, D^{i}(\mathcal F)) \to\Gamma(X, D^{i+1}(\mathcal F)))$ is not necessarily equal to ${\rm ker}(\Gamma(X, D^{i+1}(\mathcal F)) \to \Gamma(X, D^{i+2}(\mathcal F)))$ because the map $\Gamma(X, D^{i}(\mathcal F)) \to \Gamma(X, C^{i+1}(\mathcal F))$ isn't surjective. If they were equal, we would have got an exact sequence rather than just a complex.]
A: I think I have found the answers to my questions. Please verify if they're correct.
1) If we set $C^{i+1}(\mathcal{F})=D^i(\mathcal{F})/C^i(\mathcal{F})$ then it works. So, I think that's indeed what the author meant.
2) It seems that what I have said is wrong. What I have defined as a complex was in fact a co-chain complex. This is why co-homology is involved in the definition. Right?
3) $D(\mathcal{F})$ is always flabby for any sheaf $F$. Right? Functorial in this case means that $D(-)$ is a functor from the category of Abelian sheaves to the category of co-chain complexes. Is that what it means?
4) The n-th cohomology of $\Gamma(X,D^*(\mathcal{F}))$ is given by $$H^n(X,F)=\frac{\mathrm{ker}\hspace{2px}{d_{n}}}{\mathrm{Im}\hspace{2px}{d_{n-1}}}$$
where $d_n$ is the map between $\Gamma(X,D^n(\mathcal{F}))$ and $\Gamma(X,D^{n+1}(\mathcal{F}))$. Right? Therefore, since $\mathrm{Im}\hspace{2px}d_{-1} = 0$, $H^0(X,F)=\mathrm{ker}\hspace{2px}{d_{0}}=\Gamma(X,D^{0}(\mathcal{F}))$
I think there's a problem with Kempf's notation. The first index of the co-chain it has introduced is $\mathcal{F}$ and it does not agree with $D^0(\mathcal{F})$. Do you agree?
Also, about the part that it says we can put those short exact sequences together, I think it's done very trivially by pasting them together. Right?
