I am reading the book "Sheaves on Manifolds" by Kashiwara and Schapira (1990 version). (I can only find a one-page errata in https://webusers.imj-prg.fr/~pierre.schapira/BooksMono/Errata.pdf. Is there a revised version?) Up to now, I can understand most of the results and proofs. However, the proof of Proposition 2.7.5 is very confusing for me.
Why is the lower square of diagram (2.7.6) commutative? I think it is not commutative in general.
What is the middle arrow in (2.7.7)? It comes out of nowhere.
What is the final pentagon diagram? I don't believe the right arrow is correct if (2.7.6) is not commutative. And there is no explanation why the whole diagram is commutative.
Can someone explain the proof to me, or provide some reference for this type of theorem (with a more readable proof)?
After some thought, I believe the proof should be changed to the following (which is basically the same as the proof in cohomology theory in algebraic topology).
First of all, we need to add the assumption that the lower square is commutative. Otherwise, I think the Proposition is wrong.
Because $f_i=h\circ j_i$, we only need to prove $j^{\#}_0=j^{\#}_1$, i.e.,
$Rp_{X*} Rh_* h^{-1} p_X^{-1}\to Rp_{X*} Rh_* Rj_{i*} j_i^{-1} h^{-1} p_X^{-1}$
is the same for $i=0,1$.
By the commutative diagram, it is the same as the morphism
$Rp_{Y*} Rp_* p^{-1} p_Y^{-1}\to Rp_{Y*} Rp_* Rj_{i*} j_i^{-1} p^{-1} p_Y^{-1}$.
So we can try to prove the middle part
$Rp_* p^{-1}\to Rp_* Rj_{i*} j_i^{-1} p^{-1}$
is the same for j=0,1.
We can deduce from previous lemma 2.7.3 that there is a pair of isomorphism of functors $id \to Rp_* p^{-1}\to Rp_* Rj_{i*} j_i^{-1} p^{-1} = id$. In fact, the lemma implies this holds for $\{t\} \to I \to \{*\}$, and the general case $Y\times \{t\} \to Y\times I \to Y$ can be proved by considering the stalks.
In other words, the morphism $Rp_* p^{-1}\to Rp_* Rj_{i*} j_i^{-1} p^{-1}$ is the inverse of $id \to Rp_* p^{-1}$, so it is the same morphism for different $i$.