Meaning of certain cohomology classes and morphisms in Prop 3.3 of Deligne-Illusie's paper on mod $p^2$ liftings and decompos. of the de Rham complex I am still struggling with Deligne and Illusie's paper (https://eudml.org/doc/143480). They say on page 261, in the course of the proof of theorem 3.3:

The class $e(K)$ (which is associated to the degree-$1$ morphism in $H^{0}(K)\longrightarrow K\longrightarrow H^{1}(K)[-1]\longrightarrow^{+1}$, where $K$ is a decomposable complex with $K^{i}=0$ for $i\notin\left\{0,1\right\}$ over a ringed site $(T,\mathcal{O})$) is also the class of the composite morphism
\begin{equation*}
H^{0}(T,\check{N}\otimes N)\longleftarrow^{q}_{\cong}H^{1}(T,\check{N}\otimes K)\longrightarrow^{-\operatorname{Id}_{N}\otimes\operatorname{pr}}H^{2}(T,\check{N}\otimes M)
\end{equation*}
where $M=H^{0}(K)$, $N=H^{1}(K)$, and $\check{N}$ denotes the dual of $N$.
Thus
\begin{equation*}
c=(h,-g,f)\in\check{C}^{1}(U_{\bullet},\check{N}\otimes K)=\check{C}^{2}(U_{\bullet},\check{N}\otimes M)\oplus\check{C}^{1}(U_{\bullet},\check{N}\otimes K^{0})\oplus\check{C}^{0}(U_{\bullet},\check{N}\otimes K^{1})
\end{equation*}
is a $1$-cocycle of image $\operatorname{Id}_{N}$ by $q$ and, and $-h$ through $-\operatorname{Id}_{N}\otimes\operatorname{pr}$. Therefore $e(K)=-\operatorname{cl sc}(K)$.  ($U$ is a hypercovering of $T$, and $\operatorname{cl sc}$ is the class associated to the gerbe $\operatorname{sc}(K)$ of decompositions of $K$.)

Questions:
(1.) How do the morphisms $q$ and $-\operatorname{Id}_{N}\otimes\operatorname{pr}$ come about and why is $q$ a quasi-isomorphism?
(2.) How does this $c$ come about and what does it have to with everything else?
(3.) Why does that imply that $e(K)=\operatorname{cl sc}(K)$?
I promise this will be the last question concerning their paper. This paper is actually quite beautiful and easy to read, but I am having problems with chapter 3, probably because I am only superficially acquainted with sheaf cohomology on sites. Thanks in advance anyway.
 A: Let's follow the argument. first you want the inverse image of $id_N\in H^0(N\otimes\check N)=Hom(N,N)$ with the map $q$. the map $E\to N$ by definition factors through $K^1$, and $K^1\otimes\check N \to N\otimes \check N$ is surjective and first cohomology is always computed by Čech cohomology. so as usual you have to first find a cover $U$ and a lift of $id_N$ to an element of $C^0(K^1\otimes\check N)=Hom_U(N,K^1)$ which the call $f$,in now consider$$C^1(E\otimes \check N)=C^2(M\otimes\check N)\oplus C^1(K^0\otimes\check N)\oplus C^0(K^1\otimes\check N) $$(this is the usual way you compute the Čech complex of a complex,you have to first write Čech complex of each term which gives you a double complex and then you take the total complex of that double complex).
now you want a cocycle  here which has $f$ as the third component, if you write the boundary map you understand that the cocycle should be of the form $c=(h,g,f)$ where $$dg=d_0f+d_1f,h=d_0g-d_1g+d_2g.$$
the upshot is that $$e(K)=-id\otimes pr_0(q^{-1}(Id_N))=id\otimes pr_0(c)=-h.$$ now you explicitly computed $e(K)$. you can also explicitly compute the class in $H^2$ associated to a gerb(in our case sc(K)) and see that this is also $h$. in the first paragraph after proposition they give a reference for this fact:(12,IV 3.5) hence we win.
note: there is a typo in the diagram in the paper it should be $H^1(T,E\otimes \check N)$ instead of $H^1(T,K\otimes \check N)$.
