Let $X$ be a smooth Gushel–Mukai fourfold and $Y$ a smooth hyperplane section, which is a Gushel–Mukai threefold. I consider semi-orthogonal decomposition of $X$ and $Y$: $$D^b(X)=\langle\mathcal{O}_X(-H),\mathcal{U}_X,\mathcal{K}u(X),\mathcal{O}_X,\mathcal{U}^{\vee}_X\rangle$$ and $$D^b(Y)=\langle\mathcal{K}u(Y),\mathcal{O}_Y,\mathcal{U}_Y^{\vee}\rangle$$ where I denote by $\mathcal{U}_X,\mathcal{U}_Y$ the tautological bundle of $X$ and $Y$ respectively. Then I consider a functor $\operatorname{pr}\circ j_*$, where $j:Y\hookrightarrow X$ is the embedding morphism and $\operatorname{pr}$ is the projection functor, which is defined by $$\operatorname{pr}:=\operatorname{R}_{\mathcal{U}}\circ\operatorname{R}_{\mathcal{O}_X(-H)}\circ\operatorname{L}_{\mathcal{O}_X}\circ\operatorname{L}_{\mathcal{U}^{\vee}}.$$ Then we have the functor $\operatorname{pr}\circ j_*:\mathcal{K}u(Y)\rightarrow\mathcal{K}u(X)$. We can show this functor has a left adjoint functor $j^*:\mathcal{K}u(X)\rightarrow\mathcal{K}u(Y)$.
Then we prove the following proposition:
Proposition: Let $X$ be a Gushel–Mukai fourfold and $Y$ the smooth hyperplane section of $X$. Let $E\in\mathcal{K}u(Y)$ be an object, then we have a triangle $$\mathcal{S}^{-1}_{\mathcal{K}u(Y)}(E)[2]\rightarrow j^*\mathrm{pr}\circ j_*(E)\rightarrow E$$ where $\mathcal{S}_{\mathcal{K}u(Y)}$ is the Serre functor on $\mathcal{K}u(Y)$, which is nothing but $\tau\circ [2]$ with $\tau$ the involution functor. Then this triangle becomes $$\tau(E)\rightarrow j^*\operatorname{pr}\circ j_*(E)\rightarrow E.$$ Now we want to show that if $E$ is a stable object in $\mathcal{K}u(Y)$ with respect to a Serre-invariant stability condition and $E'$ is another stable object (with respect to the same stability condition) such that $[E']=[E]\in\mathcal{N}(\mathcal{K}u(Y))$, where $\mathcal{N}(\mathcal{K}u(Y))$ is the numerical Grothendieck group of $\mathcal{K}u(Y)$. Then $\operatorname{Hom}(\operatorname{pr}\circ j_*(E),\operatorname{pr}\circ j_*(E'))=0$. To show this, we apply $\operatorname{Hom}({-},E')$ the above triangle, then we get a long exact sequence: $$\operatorname{Hom}(E',E)\rightarrow\operatorname{Hom}(j^*\operatorname{pr}\circ j_*(E),E')\rightarrow\operatorname{Hom}(\tau(E),E')\xrightarrow{\alpha}\operatorname{Ext}^1(E,E')\rightarrow\dotsb.$$
Note that $\operatorname{Hom}(E',E)=0$ since $E$, $E'$ are both stable objects with the same character and they are not isomorphic, then by Schur's Lemma, the Hom vanishes. Now if $\tau(E)\neq E'$, then by Schur's Lemma again, we get $\operatorname{Hom}(\tau(E),E')=0$ since $\tau$ does not change the character and the stability of $E$. Then we are done. But if $\tau(E)\cong E'$, this $\operatorname{Hom}(\tau(E),E')=k$, so we have to look at the next term $\operatorname{Ext}^1(E,E')=\operatorname{Hom}(E,E'[1])=\operatorname{Hom}(E'[1],\tau(E)[2])=\operatorname{Hom}(E',E'[1])$. Then we need to look at the map $$\operatorname{Hom}(\tau(E),E')=\operatorname{Hom}(E',E')\xrightarrow{\alpha}\operatorname{Hom}(E',E'[1]).$$ We can show $\operatorname{Hom}(E',E'[1])$ is always a vector space of dimension greater than one, so we only need to show the map $\alpha$ is non-trivial, as a result, $0=\operatorname{Ker}(\alpha)=\operatorname{Hom}(j^*\operatorname{pr}\circ j_*(E),E')\cong\operatorname{Hom}(\operatorname{pr}\circ j_*(E),\operatorname{pr}\circ j_*(E'))$.
My question is whether the map $\alpha$ is non-trivial? Note that this map is just compose the identity element in $\operatorname{Hom}(E',E')$ with an element $\gamma\in\operatorname{Hom}(E',E'[1])$, i.e. $\alpha(\operatorname{id})=\operatorname{id}\circ\gamma$.
