Proof of Birger Iversen "Cohomology of Sheaves" Theorem 6.8 I am having troubles completing the proof of theorem 6.8 (page 44) from Birger Iversen, Cohomology of Sheaves. (pdf here)
Previously we had constructed a functor $\rho$ from $K^+(A)$ (the homotopy category of bounded below complexes in $A$) to $D^+$ (The homotopy category of bounded below complexes of injectives in $A$).
The theorem says that $\rho$ transforms triangles into triangles. The last line of the proof says "It is easy to conclude the proof by means of 6.2", but i can't figure out how to do it. This is what I got so far:
$$\begin{array} $\rho X^\circ& \longrightarrow&\rho Y^\circ& \longrightarrow &Con^\circ(\rho f)& \longrightarrow& \rho X^\circ [1] \\
\downarrow{1}&&\downarrow{1}&&\downarrow{[\phi,1]^{-1}c}&&\downarrow{1}&&\\
\rho X^\circ& \longrightarrow&\rho Y^\circ& \longrightarrow &\rho Z^\circ& \longrightarrow& \rho X^\circ [1] \\
\end{array}
$$
Where $\phi$ is the arrow that comes from "filling in the third arrow" (see proof of iversen). In order to prove that the lower line is a triangle we have to show that this diagram is homotopy commutative (which I managed to do) and that the vertical arrows are homotopy equivalences. 
So my question is: "How do you prove that $[\phi,1]^{-1} c $ is a homotopy equivalence ?"
Or finishing the proof in another way would also help me, thanks in advance.
 A: I just saw the question. Sorry if this is too late. 
One has that $\phi:Z^\bullet\to Con^\bullet\left( \rho f\right)$ is a quasi-isomorphism, and that $c:Z^\bullet\to \rho Z^\bullet$ is a quasi-isomorphism. Observe that $Con^\bullet\left( \rho f\right)$ is injective (it's the coproduct of injectives). By Theorem 6.2 $[c,1]:[\rho Z^\bullet, Con^\bullet\left(\rho f\right)]\to [Z^\bullet,Con^\bullet\left(\rho f\right)]$ and $[\phi,1]:[Con^\bullet\left(\rho f\right),\rho Z^\bullet]\to [Z^\bullet,\rho Z^\bullet]$ are isomorphisms. These isomorphisms imply the existence and uniqueness up to homotopy of $\xi:\rho Z^\bullet\to Con^\bullet\left( \rho f\right)$ and $\xi^\prime:Con^\bullet\left(\rho f\right)\to \rho Z^\bullet$ such that $\xi \circ c \simeq \phi$ and $\xi^\prime\circ \phi \simeq c.$
In particular for $\xi\circ\xi^\prime:Con^\bullet\left( \rho f\right)\to Con^\bullet\left( \rho f\right)$ and $\xi^\prime\circ\xi:\rho Z^\bullet\to \rho Z^\bullet,$ we obtain:
\begin{align*}
\xi^\prime \circ \xi \circ c \simeq c,\\
\xi\circ \xi^\prime\circ \phi \simeq \phi. 
\end{align*}
The latter is the same as
\begin{align*}
[c,1]\left( \xi^\prime\circ\xi\right) &=[c,1]\left({1}_{\rho Z^\bullet}\right),\\
[\phi,1]\left( \xi\circ\xi^\prime\right) &=[\phi,1]\left({1}_{Con^\bullet\left( \rho f\right)}\right). 
\end{align*} 
Since these maps are isomorphisms of the homotopy classes, it follows that $\xi$ and $\xi^\prime$ are homotopy equivalences. The homotopy commutativity of your diagram follows from an analogous reasoning ($[\rho Y^\bullet, Con^\bullet(\rho f)]\cong [Y^\bullet,Con^\bullet(\rho f)]$ by $[c,1]$ and the corresponding case for $X^\bullet$).
Hope this helps!
