This question is based on Demailly's note, Complex Analytic and Differential Geometry, IV.13.8, p.233.
$F: X \rightarrow Y$ is continuous and $\mathscr{A}$ is a sheaf of abelian groups on $X$. For the Godement resolution $0\rightarrow \mathscr{A} \rightarrow \mathscr{A}^{[0]} \rightarrow \mathscr{A}^{[1]} \rightarrow \cdots$, we can associate a double complex $K^{p,q} = (F_\star(A^{[p]}))^{[q]}$. Consider the Leray spectral sequence of $F$. The edge homomorphism $$H^l(Y, F_\star\mathscr{A}) \twoheadrightarrow E^{l,0}_\infty \hookrightarrow H^l(K^{[\bullet,\bullet]}) \cong H^l(F_\star(\mathscr{A}^{[\bullet]}))= H^l(X,\mathscr{A})$$ is not obviously canonical (This 'edge homomorphism' is actually a composite of two edge homomorphisms of two different spectral sequence). Now for short, you can just jump down to the bottom of the question.
However, in Demailly's note, Complex Analytic and Differential Geometry, IV.13.8, p.233, he gives that the edge homomorphism coninsides with the composite morphism $$F^\#: H^l(Y,F_\star\mathscr{A}) \overset{F^\star}{\longrightarrow} H^l(X,F^{-1}F_\star \mathscr{A})\overset{H^l(\mu_F)}{\longrightarrow} H^l(X,\mathscr{A})$$ where $\mu_F : F^{-1}F_\star \mathscr{A} \longrightarrow \mathscr{A}$ is the canonical sheaf morphism.
In the proof he at first construction a commutative diagram $$\require{AMScd} \begin{CD} H^l(X,\mathscr{A}) @<<< H^l(Y,F_\star\mathscr{A})\\ @AF^\#AA @AA\mathrm{Id}A\\ H^l(Y,F_\star\mathscr{A}) @<<< H^l(Y,F_\star\mathscr{A}) \end{CD} $$ where the upper and lower arrows are edge homomorphisms. Then we need only prove the situation where $F = \mathrm{Id}$. And then Demailly says this is an immediate consequence of the fact that we have a quasi-isomorphism $$(\cdots \rightarrow 0 \rightarrow \mathscr{A} \rightarrow 0 \rightarrow \cdots) \longrightarrow \mathscr{A}^{[\bullet]}$$ where $\mathscr{A}^{[\bullet]}$ is the Godement flabby resolution of $\mathscr{A}$.
My question is why the quasi-isomorphism of resolution implies that the edge homomorphism is identity?
Just consider $F=\mathrm{Id}$. Let $\{E_r^{p,q}\}$ be the spectral sequence associated to filtration $K_p=\otimes_{l \ge p}K^{[l,\bullet]}$, and $\{ \tilde{E}_r^{p,q} \}$ be the spectral sequence associated to the filtration $\tilde{K}_p = \otimes_{l \ge p}K^{[\bullet , l]}$. The edge homomorphism constructed first maps $H^l(X,\mathscr{A}) = E_2^{l,0}$ into $H^l(K^{[\bullet,\bullet]})$, then maps out from $H^l(K^{[\bullet,\bullet]})$ into $\tilde{E}_2^{0,l} = H^l(X,\mathscr{A})$. What the theorem assert is that this morphism is actually identity! That's what I am confused about.
This question has an answer in Natural morphism appearing in Grothendieck spectral sequence. The answer involves some categorical and homologicalll algebraic argument.