Lemma 57.11.3. Let $F$ and $F'$ be siblings as in Definition 57.11.1. Then

1. if $F$ is essentially surjective, then $F'$ is essentially surjective,

2. if $F$ is fully faithful, then $F'$ is fully faithful.

Proof. Part (1) is immediate from property (2) for siblings.

Assume $F$ is fully faithful. Denote $\mathcal{D}' \subset \mathcal{D}$ the essential image of $F$ so that $F : D^ b(\mathcal{A}) \to \mathcal{D}'$ is an equivalence. Since the functor $F'$ factors through $\mathcal{D}'$ by property (2) for siblings, we can consider the functor $H = F^{-1} \circ F' : D^ b(\mathcal{A}) \to D^ b(\mathcal{A})$. Observe that $H$ is a sibling of the identity functor. Since it suffices to prove that $H$ is fully faithful, we reduce to the problem discussed in the next paragraph.

Set $\mathcal{D} = D^ b(\mathcal{A})$. We have to show a sibling $F : \mathcal{D} \to \mathcal{D}$ of the identity functor is fully faithful. Denote $a_ X : X \to F(X)$ the functorial isomorphism for $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ given to us by Definition 57.11.1. For any $K$ in $\mathcal{D}$ and distinguished triangle $K_1 \to K_2 \to K_3$ of $\mathcal{D}$ if the maps

$F : \mathop{\mathrm{Hom}}\nolimits (K, K_ i[n]) \to \mathop{\mathrm{Hom}}\nolimits (F(K), F(K_ i[n]))$

are isomorphisms for all $n \in \mathbf{Z}$ and $i = 1, 3$, then the same is true for $i = 2$ and all $n \in \mathbf{Z}$. This uses the $5$-lemma Homology, Lemma 12.5.20 and Derived Categories, Lemma 13.4.2; details omitted. Similarly, if the maps

$F : \mathop{\mathrm{Hom}}\nolimits (K_ i[n], K) \to \mathop{\mathrm{Hom}}\nolimits (F(K_ i[n]), F(K))$

are isomorphisms for all $n \in \mathbf{Z}$ and $i = 1, 3$, then the same is true for $i = 2$ and all $n \in \mathbf{Z}$. Using the canonical truncations and induction on the number of nonzero cohomology objects, we see that it is enough to show

$F : \mathop{\mathrm{Ext}}\nolimits ^ q(X, Y) \to \mathop{\mathrm{Ext}}\nolimits ^ q(F(X), F(Y))$

is bijective for all $X, Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ and all $q \in \mathbf{Z}$. Since $F$ is a sibling of $\text{id}$ we have $F(X) \cong X$ and $F(Y) \cong Y$ hence the right hand side is zero for $q < 0$. The case $q = 0$ is OK by our assumption that $F$ is a sibling of the identity functor. It remains to prove the cases $q > 0$.

The case $q = 1$: Injectivity. An element $\xi$ of $\mathop{\mathrm{Ext}}\nolimits ^1(X, Y)$ gives rise to a distinguished triangle

$Y \to E \to X \xrightarrow {\xi } Y[1]$

Observe that $E \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. Since $F$ is a sibling of the identity functor we obtain a commutative diagram

$\xymatrix{ E \ar[d] \ar[r] & X \ar[d] \\ F(E) \ar[r] & F(X) }$

whose vertical arrows are the isomorphisms $a_ E$ and $a_ X$. By TR3 the distinguished triangle associated to $\xi$ we started with is isomorphic to the distinguished triangle

$F(Y) \to F(E) \to F(X) \xrightarrow {F(\xi )} F(Y[1]) = F(Y)[1]$

Thus $\xi = 0$ if and only if $F(\xi )$ is zero, i.e., we see that $F : \mathop{\mathrm{Ext}}\nolimits ^1(X, Y) \to \mathop{\mathrm{Ext}}\nolimits ^1(F(X), F(Y))$ is injective.

The case $q = 1$: Surjectivity. Let $\theta$ be an element of $\mathop{\mathrm{Ext}}\nolimits ^1(F(X), F(Y))$. This defines an extension of $F(X)$ by $F(Y)$ in $\mathcal{A}$ which we may write as $F(E)$ as $F$ is a sibling of the identity functor. We thus get a distinguished triangle

$F(Y) \xrightarrow {F(\alpha )} F(E) \xrightarrow {F(\beta )} F(X) \xrightarrow {\theta } F(Y[1]) = F(Y)[1]$

for some morphisms $\alpha : Y \to E$ and $\beta : E \to X$. Since $F$ is a sibling of the identity functor, the sequence $0 \to Y \to E \to X \to 0$ is a short exact sequence in $\mathcal{A}$! Hence we obtain a distinguished triangle

$Y \xrightarrow {\alpha } E \xrightarrow {\beta } X \xrightarrow {\delta } Y[1]$

for some morphism $\delta : X \to Y[1]$. Applying the exact functor $F$ we obtain the distinguished triangle

$F(Y) \xrightarrow {F(\alpha )} F(E) \xrightarrow {F(\beta )} F(X) \xrightarrow {F(\delta )} F(Y)[1]$

Arguing as above, we see that these triangles are isomorphic. Hence there exists a commutative diagram

$\xymatrix{ F(X) \ar[d]^\gamma \ar[r]_{F(\delta )} & F(Y[1]) \ar[d]_\epsilon \\ F(X) \ar[r]^\theta & F(Y[1]) }$

for some isomorphisms $\gamma$, $\epsilon$ (we can say more but we won't need more information). We may write $\gamma = F(\gamma ')$ and $\epsilon = F(\epsilon ')$. Then we have $\theta = F(\epsilon ' \circ \delta \circ (\gamma ')^{-1})$ and we see the surjectivity holds.

The case $q > 1$: surjectivity. Using Yoneda extensions, see Derived Categories, Section 13.27, we find that for any element $\xi$ in $\mathop{\mathrm{Ext}}\nolimits ^ q(F(X), F(Y))$ we can find $F(X) = B_0, B_1, \ldots , B_{q - 1}, B_ q = F(Y) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ and elements

$\xi _ i \in \mathop{\mathrm{Ext}}\nolimits ^1(B_{i - 1}, B_ i)$

such that $\xi$ is the composition $\xi _ q \circ \ldots \circ \xi _1$. Write $B_ i = F(A_ i)$ (of course we have $A_ i = B_ i$ but we don't need to use this) so that

$\xi _ i = F(\eta _ i) \in \mathop{\mathrm{Ext}}\nolimits ^1(F(A_{i - 1}), F(A_ i)) \quad \text{with}\quad \eta _ i \in \mathop{\mathrm{Ext}}\nolimits ^1(A_{i - 1}, A_ i)$

by surjectivity for $q = 1$. Then $\eta = \eta _ q \circ \ldots \circ \eta _1$ is an element of $\mathop{\mathrm{Ext}}\nolimits ^ q(X, Y)$ with $F(\eta ) = \xi$.

The case $q > 1$: injectivity. An element $\xi$ of $\mathop{\mathrm{Ext}}\nolimits ^ q(X, Y)$ gives rise to a distinguished triangle

$Y[q - 1] \to E \to X \xrightarrow {\xi } Y[q]$

Applying $F$ we obtain a distinguished triangle

$F(Y)[q - 1] \to F(E) \to F(X) \xrightarrow {F(\xi )} F(Y)[q]$

If $F(\xi ) = 0$, then $F(E) \cong F(Y)[q - 1] \oplus F(X)$ in $\mathcal{D}$, see Derived Categories, Lemma 13.4.11. Since $F$ is a sibling of the identity functor we have $E \cong F(E)$ and hence

$E \cong F(E) \cong F(Y)[q - 1] \oplus F(X) \cong Y[q - 1] \oplus X$

In other words, $E$ is isomorphic to the direct sum of its cohomology objects. This implies that the initial distinguished triangle is split, i.e., $\xi = 0$. $\square$

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