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A bit more justification is needed in the proof as to why the two $C:=B \otimes_k L^{op}$-module structures of $M$ coincide, and I propose the following. Call these two $C$-modules $M_1$ and $M_2$. They have the same $k$-dimension which is the $k$-dimension of $M$. Each of them is a direct sum of simple $C$-modules. As there is up to $C$-isomorphism a unique simple $C$-module, $M_1$ and $M_2$ are isomorphic as $C$-modules. This isomorphism induces an $L$-isomorphism of $M$, which must be multiplication by a unit of $A$.

I would change $d_1+\dots+d_s+1$ to either $d_1+\dots + d_s$ (the cleanest looking) or $(d_1-1)+\dots+(d_s-1)+1$ (the sharp one). Hopefully I am not making a mistake.

Sorry my last comment is wrong. Please ignore it. The typo would be "$\varphi : f^{-1}\mathcal{O}_ {S'}[\mathcal{E}] \to \mathcal{O}_ {X'}$" the definition of the ideal $\mathcal{I}'=\mathrm{ker}(f^{-1}\mathcal{O}_ {S'}[\mathcal{E}]\rightarrow \mathcal{O}_ X$).

Typo : First line of the thrid paragraph : "$\varphi:\mathcal{O}_ {S'}[\mathcal{E}]\rightarrow \mathcal{O}_X$".

Sorry, it doesn't seem to like #'s in math mode so I'll write * instead. Cayley-Hamilton says the polynomial $s^*(P) \in B[x]$ annihilates the element $t^*(f) \in B$. Since the coefficients of $P$ are in $C$, we get that $t^*(P(f)) = 0$ in $B$. Since $t: \operatorname{Spec}(B) \to \operatorname{Spec}(A)$ is faithfully flat, we get that $P(f) = 0$.

This is maybe too much detail but just to spell out the last sentence: Cayley-Hamilton says the polynomial $s^#(P) \in B[x]$ annihilates $t^#(f) \in B$. Since the coefficients of $P$ are in $C$, we get $t^#(P(x))$ which implies $P(x) = 0$ as $t$ is faithfully flat.

Another typo (sorry...) : before the defition of $a$ and $b$ : "we obtain two homomorphisms of $\mathcal{O}_{S_1'}$-algebras..."

It seems that in step $3$, the $\mathcal{O}_{S_ 1 '}$-algebra morphism $b\pm \theta\circ d :\mathcal{O}_{S_ 1'}[\mathcal{E}]\rightarrow \mathcal{O}_{X_ 2'}$ will agree with $\nu \circ a: \mathcal{I}'\rightarrow \mathcal{G}_ 2$ over $\mathcal{I}'$ instead of being $0$. (In fact in the proof, it is constructed $\mathcal{O}_{S_ 1'}[\mathcal{E}]/\mathcal{I}'=\mathcal{O}_{X_ 1}$ rather than $\mathcal{O}_{X_ 1'}$. ) One may still conclude step $3$ as in the proof of Lemma 91.2.1 by showing that $\mathcal{O}_{X_ 1'}$ is the pushout of $\mathcal{I}'\rightarrow \mathcal{O}_{S_1 '}[\mathcal{E}]$ and $\mathcal{I}'\rightarrow \mathcal{G}_ 1$.

Typo : in the definition of $\xi$ : "...which sends the class of a local section $f$ of $\mathcal{I}$ to $\nu(a(f'))-b(f')$..."

Typo : $I_1A_1'[E]$ is written as $I_1A_1[E]$ twice in the proof: "Note that there is a surjection $J'\rightarrow J$ whose kernel is $I_1A_1'[ E ]$ " and "...we see that these maps agree on $I_1A_1'[E]$.

Also a silly question : could you spill out how the commutative square involing $c_1$ and $c_2$ implies that $a$ and $b$ agree on $I_1A_1'[E]$ ? I tried some diagram chasing with no success.

In (3) of the statement, "equivalence" should say "equivalence relation."

My bad, the links didn't worked. The first one is Karmazyn--Kuznetsov--Shinder "Derived categories of singular surfaces" (arXiv:1809.10628), and the second one is Mauri--Shinder "Homological Bondal-Orlov localization conjecture for rational singularities" (arXiv:2212.06786)

Say for example $f\colon \tilde{X} \to X$ is a resolution of a variety with rational singularties. There is a semi-orthogonal decomposition $D^-(\tilde{X}) = \langle \ker R\pi_\ast, L\pi^\ast D^-(X) \rangle$, and so $R\pi_\ast\colon D^-(\tilde{X}) \to D^-(X)$ is a Verdier quotient on the left unbounded categories. One could ask if this is true also for the bounded derived categories, in which case the criterion can be used. See for instance the proof of Proposition 2.19 (pp. 17--8) of \ref{https://arxiv.org/pdf/1809.10628}, and the introduction of \ref{https://arxiv.org/pdf/2212.06786}.

Minor typos: every instance of $M$ in $\cdots \otimes^{\mathbf{L}} Rf_* M$ and $\cdots \otimes^{\mathbf{L}} M$ should be replaced by $K$.

$L$ is not used in the remark.

If we want to stay consistent with the previous notations, we should change all instances like $\mathrm{Hom}(K, M) \otimes M$ to $\mathrm{Hom}(L, K) \otimes L$, update the bottom right corner to $Rf_* K$ and remove $M$ in "let $K$, $L$, $M$ be objects of ...$.

Thanks and fixed here.

Okay, I did this here.

Before you do: what is an example where one might use that?

Fixed here.

Indeed. Fixed here.

Your argument uses results from chapters beyond this one, so it won't work. You can probably dumb down the proof quite a bit, but I actually find it quite an amusing argument.