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Dag Oskar Madsen
Dag Oskar Madsen
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Here are some more details.

Lemma 3.2.4 written in the notation of Theorem 3.2.5 becomes the statement that $$\phi_{M,N[n]}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n])$$ is an isomorphism for all $M,N$ in $\mathcal Y(T)$ and $n \geq 0$. It is also an isomorphism when $n<0$ since then both sides are $0$.

For any $N$ in $\operatorname{mod} \Lambda$ there is a triangle $$N'\to P\to N\to N'[1],$$$$N'\to P\to N\to N'[1]$$ with $N',P$ in $\mathcal Y(T)$, and for all $n \in \mathbb Z$ we get a commutative diagram $\require{AMScd}$ \begin{CD} \scriptsize {\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(M,N'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,P[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N'[n+1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,P[n+1])\\ @VV{\wr}V @VV{\wr}V @VVV @VV{\wr}V @VV{\wr}V\\ \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TP[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n]) @>>> \scriptsize {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN'[n+1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TP[n+1]) \end{CD}

Since all the other downward arrows are isomorphisms, it follows from the five lemma that the middle downward arrow is an isomorphism. So $$\phi_{M,N[n]}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n])$$ is an isomorphism for all $M$ in $\mathcal Y(T)$, $N$ in $\operatorname{mod} \Lambda$ and $n \in \mathbb Z$. A similar argument in the other variable gives that $\phi_{M,N[n]}$ is an isomorphism for all $M,N$ in $\operatorname{mod} \Lambda$ and $n \in \mathbb Z$.

This is the basis for the induction. The inductive step uses triangles of the form $$Y' \to Y \to Y'' \to Y'[1]$$ with $\ell(Y') = \ell(Y)-1$ and $\ell(Y'') = 1$. We apply the five lemma to the commutative diagram $\require{AMScd}$ \begin{CD} \scriptsize {\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y''[n-1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y''[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y'[n+1])\\ @VV{\wr}V @VV{\wr}V @VVV @VV{\wr}V @VV{\wr}V\\ \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY''[n-1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY[n]) @>>> \scriptsize {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY''[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY'[n+1]) \end{CD}

and a similar version for the other variable to conclude that $$\phi_{X,Y}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(X,Y) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY)$$ is an isomorphism for all $X,Y$ in ${\mathbf D^b(\operatorname{mod} \Lambda)}$.

Here are some more details.

Lemma 3.2.4 written in the notation of Theorem 3.2.5 becomes the statement that $$\phi_{M,N[n]}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n])$$ is an isomorphism for all $M,N$ in $\mathcal Y(T)$ and $n \geq 0$. It is also an isomorphism when $n<0$ since then both sides are $0$.

For any $N$ in $\operatorname{mod} \Lambda$ there is a triangle $$N'\to P\to N\to N'[1],$$ and for all $n \in \mathbb Z$ we get a commutative diagram $\require{AMScd}$ \begin{CD} \scriptsize {\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(M,N'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,P[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N'[n+1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,P[n+1])\\ @VV{\wr}V @VV{\wr}V @VVV @VV{\wr}V @VV{\wr}V\\ \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TP[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n]) @>>> \scriptsize {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN'[n+1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TP[n+1]) \end{CD}

Since all the other downward arrows are isomorphisms, it follows from the five lemma that the middle downward arrow is an isomorphism. So $$\phi_{M,N[n]}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n])$$ is an isomorphism for all $M$ in $\mathcal Y(T)$, $N$ in $\operatorname{mod} \Lambda$ and $n \in \mathbb Z$. A similar argument in the other variable gives that $\phi_{M,N[n]}$ is an isomorphism for all $M,N$ in $\operatorname{mod} \Lambda$ and $n \in \mathbb Z$.

This is the basis for the induction. The inductive step uses triangles of the form $$Y' \to Y \to Y'' \to Y'[1]$$ with $\ell(Y') = \ell(Y)-1$ and $\ell(Y'') = 1$. We apply the five lemma to the commutative diagram $\require{AMScd}$ \begin{CD} \scriptsize {\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y''[n-1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y''[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y'[n+1])\\ @VV{\wr}V @VV{\wr}V @VVV @VV{\wr}V @VV{\wr}V\\ \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY''[n-1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY[n]) @>>> \scriptsize {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY''[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY'[n+1]) \end{CD}

and a similar version for the other variable to conclude that $$\phi_{X,Y}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(X,Y) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY)$$ is an isomorphism for all $X,Y$ in ${\mathbf D^b(\operatorname{mod} \Lambda)}$.

Here are some more details.

Lemma 3.2.4 written in the notation of Theorem 3.2.5 becomes the statement that $$\phi_{M,N[n]}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n])$$ is an isomorphism for all $M,N$ in $\mathcal Y(T)$ and $n \geq 0$. It is also an isomorphism when $n<0$ since then both sides are $0$.

For any $N$ in $\operatorname{mod} \Lambda$ there is a triangle $$N'\to P\to N\to N'[1]$$ with $N',P$ in $\mathcal Y(T)$, and for all $n \in \mathbb Z$ we get a commutative diagram $\require{AMScd}$ \begin{CD} \scriptsize {\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(M,N'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,P[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N'[n+1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,P[n+1])\\ @VV{\wr}V @VV{\wr}V @VVV @VV{\wr}V @VV{\wr}V\\ \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TP[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n]) @>>> \scriptsize {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN'[n+1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TP[n+1]) \end{CD}

Since all the other downward arrows are isomorphisms, it follows from the five lemma that the middle downward arrow is an isomorphism. So $$\phi_{M,N[n]}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n])$$ is an isomorphism for all $M$ in $\mathcal Y(T)$, $N$ in $\operatorname{mod} \Lambda$ and $n \in \mathbb Z$. A similar argument in the other variable gives that $\phi_{M,N[n]}$ is an isomorphism for all $M,N$ in $\operatorname{mod} \Lambda$ and $n \in \mathbb Z$.

This is the basis for the induction. The inductive step uses triangles of the form $$Y' \to Y \to Y'' \to Y'[1]$$ with $\ell(Y') = \ell(Y)-1$ and $\ell(Y'') = 1$. We apply the five lemma to the commutative diagram $\require{AMScd}$ \begin{CD} \scriptsize {\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y''[n-1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y''[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y'[n+1])\\ @VV{\wr}V @VV{\wr}V @VVV @VV{\wr}V @VV{\wr}V\\ \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY''[n-1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY[n]) @>>> \scriptsize {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY''[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY'[n+1]) \end{CD}

and a similar version for the other variable to conclude that $$\phi_{X,Y}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(X,Y) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY)$$ is an isomorphism for all $X,Y$ in ${\mathbf D^b(\operatorname{mod} \Lambda)}$.

Source Link
Dag Oskar Madsen
Dag Oskar Madsen
  • 3.7k
  • 3
  • 28
  • 51

Here are some more details.

Lemma 3.2.4 written in the notation of Theorem 3.2.5 becomes the statement that $$\phi_{M,N[n]}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n])$$ is an isomorphism for all $M,N$ in $\mathcal Y(T)$ and $n \geq 0$. It is also an isomorphism when $n<0$ since then both sides are $0$.

For any $N$ in $\operatorname{mod} \Lambda$ there is a triangle $$N'\to P\to N\to N'[1],$$ and for all $n \in \mathbb Z$ we get a commutative diagram $\require{AMScd}$ \begin{CD} \scriptsize {\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(M,N'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,P[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N'[n+1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,P[n+1])\\ @VV{\wr}V @VV{\wr}V @VVV @VV{\wr}V @VV{\wr}V\\ \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TP[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n]) @>>> \scriptsize {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN'[n+1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TP[n+1]) \end{CD}

Since all the other downward arrows are isomorphisms, it follows from the five lemma that the middle downward arrow is an isomorphism. So $$\phi_{M,N[n]}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n])$$ is an isomorphism for all $M$ in $\mathcal Y(T)$, $N$ in $\operatorname{mod} \Lambda$ and $n \in \mathbb Z$. A similar argument in the other variable gives that $\phi_{M,N[n]}$ is an isomorphism for all $M,N$ in $\operatorname{mod} \Lambda$ and $n \in \mathbb Z$.

This is the basis for the induction. The inductive step uses triangles of the form $$Y' \to Y \to Y'' \to Y'[1]$$ with $\ell(Y') = \ell(Y)-1$ and $\ell(Y'') = 1$. We apply the five lemma to the commutative diagram $\require{AMScd}$ \begin{CD} \scriptsize {\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y''[n-1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y''[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y'[n+1])\\ @VV{\wr}V @VV{\wr}V @VVV @VV{\wr}V @VV{\wr}V\\ \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY''[n-1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY[n]) @>>> \scriptsize {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY''[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY'[n+1]) \end{CD}

and a similar version for the other variable to conclude that $$\phi_{X,Y}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(X,Y) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY)$$ is an isomorphism for all $X,Y$ in ${\mathbf D^b(\operatorname{mod} \Lambda)}$.