hcju/stacksqa huggingface dataset:
By Lemma \ref{lemma-2-out-of-3-defined} we know that $RF$ is defined at $Z$ and that $RF$ applied to the triangle produces a distinguished triangle. Consider the morphism of distinguished triangles $$ \xymatrix{ (F(X), F(Y), F(Z), F(f), F(g), F(h)) \ar[d] \\ (RF(X), RF(Y), RF(Z), RF(f), RF(g), RF(h)) } $$ Two out of three maps are isomorphisms, hence so is the third.
If $X \oplus Y$ computes $RF$, then $RF(X \oplus Y) = F(X) \oplus F(Y)$. In the proof of Lemma \ref{lemma-direct-sum-defined} we have seen that the functor $X/S \times Y/S \to (X \oplus Y)/S$, $(s, s') \mapsto s \oplus s'$ is cofinal. We will use this without further mention. Let $s : X \to X'$ be an element of $S$. Then $F(X) \to F(X')$ has a section, namely, $$ F(X') \to F(X' \oplus Y) \to RF(X' \oplus Y) = RF(X \oplus Y) = F(X) \oplus F(Y) \to F(X). $$ where we have used Lemma \ref{lemma-derived-inverts}. Hence $F(X') = F(X) \oplus E$ for some object $E$ of $\mathcal{D}'$ such that $E \to F(X' \oplus Y) \to RF(X'\oplus Y) = RF(X \oplus Y)$ is zero (Lemma \ref{lemma-when-split}). Because $RF$ is defined at $X' \oplus Y$ with value $F(X) \oplus F(Y)$ we can find a morphism $t : X' \oplus Y \to Z$ of $S$ such that $F(t)$ annihilates $E$. We may assume $Z = X'' \oplus Y''$ and $t = t' \oplus t''$ with $t', t'' \in S$. Then $F(t')$ annihilates $E$. It follows that $F$ is essentially constant on $X/S$ with value $F(X)$ as desired.
In this proof we try to be careful. Hence let us think of the derived functors as the functors $$ RF' : S^{-1}\mathcal{A} \to (S')^{-1}\mathcal{B}, \quad R(G \circ F) : S^{-1}\mathcal{A} \to \mathcal{C}, \quad RG : (S')^{-1}\mathcal{B} \to \mathcal{C}. $$ Let us denote $Q_A : \mathcal{A} \to S^{-1}\mathcal{A}$ and $Q_B : \mathcal{B} \to (S')^{-1}\mathcal{B}$ the localization functors. Then $F' = Q_B \circ F$. Note that for every object $Y$ of $\mathcal{B}$ there is a canonical map $$ G(Y) \longrightarrow RG(Q_B(Y)) $$ in other words, there is a transformation of functors $t' : G \to RG \circ Q_B$. Let $X$ be an object of $\mathcal{A}$. We have \begin{align*} R(G \circ F)(Q_A(X)) & = \colim_{s : X \to X' \in S} G(F(X')) \\ & \xrightarrow{t'} \colim_{s : X \to X' \in S} RG(Q_B(F(X'))) \\ & = \colim_{s : X \to X' \in S} RG(F'(X')) \\ & = RG(\colim_{s : X \to X' \in S} F'(X')) \\ & = RG(RF'(X)). \end{align*} The system $F'(X')$ is essentially constant in the category $(S')^{-1}\mathcal{B}$. Hence we may pull the colimit inside the functor $RG$ in the third equality of the diagram above, see Categories, Lemma \ref{categories-lemma-image-essentially-constant} and its proof. We omit the proof this defines a transformation of functors. The case of left derived functors is similar.
Let $X$ be an object of $K^{+}(\mathcal{A})$. Consider a quasi-isomorphism $s : X \to X'$ in $K(\mathcal{A})$. By Lemma \ref{lemma-complex-cohomology-bounded} there exists quasi-isomorphism $X' \to X''$ with $X''$ bounded below. Hence we see that $X/\text{Qis}^+(\mathcal{A})$ is cofinal in $X/\text{Qis}(\mathcal{A})$. Thus it is clear that (1) holds. Part (2) follows directly from part (1). Parts (3) and (4) are dual to parts (1) and (2).
We may add $0$ to $\mathcal{I}$ if necessary. Pick $A \in \mathcal{I}$. Let $A[0] \to K^\bullet$ be a quasi-isomorphism with $K^\bullet$ bounded below. Then we can find a quasi-isomorphism $K^\bullet \to I^\bullet$ with $I^\bullet$ bounded below and each $I^n \in \mathcal{I}$, see Lemma \ref{lemma-subcategory-right-resolution}. Hence we see that these resolutions are cofinal in the category $A[0]/\text{Qis}^{+}(\mathcal{A})$. To finish the proof it therefore suffices to show that for any quasi-isomorphism $A[0] \to I^\bullet$ with $I^\bullet$ bounded above and $I^n \in \mathcal{I}$ we have $F(A)[0] \to F(I^\bullet)$ is a quasi-isomorphism. To see this suppose that $I^n = 0$ for $n < n_0$. Of course we may assume that $n_0 < 0$. Starting with $n = n_0$ we prove inductively that $\Im(d^{n - 1}) = \Ker(d^n)$ and $\Im(d^{-1})$ are elements of $\mathcal{I}$ using property (2) and the exact sequences $$ 0 \to \Ker(d^n) \to I^n \to \Im(d^n) \to 0. $$ Moreover, property (2) also guarantees that the complex $$ 0 \to F(I^{n_0}) \to F(I^{n_0 + 1}) \to \ldots \to F(I^{-1}) \to F(\Im(d^{-1})) \to 0 $$ is exact. The exact sequence $0 \to \Im(d^{-1}) \to I^0 \to I^0/\Im(d^{-1}) \to 0$ implies that $I^0/\Im(d^{-1})$ is an element of $\mathcal{I}$. The exact sequence $0 \to A \to I^0/\Im(d^{-1}) \to \Im(d^0) \to 0$ then implies that $\Im(d^0) = \Ker(d^1)$ is an elements of $\mathcal{I}$ and from then on one continues as before to show that $\Im(d^{n - 1}) = \Ker(d^n)$ is an element of $\mathcal{I}$ for all $n > 0$. Applying $F$ to each of the short exact sequences mentioned above and using (2) we observe that $F(A)[0] \to F(I^\bullet)$ is an isomorphism as desired.
Assume $K^\bullet$ satisfies the assumptions of (1). Let $K^\bullet \to L^\bullet$ be any quasi-isomorphism. Then it is also true that $K^\bullet \to \tau_{\geq a}L^\bullet$ is a quasi-isomorphism by our assumption on $K^\bullet$. Hence in the category $K^\bullet/\text{Qis}^{+}(\mathcal{A})$ the quasi-isomorphisms $s : K^\bullet \to L^\bullet$ with $L^n = 0$ for $n < a$ are cofinal. Thus $RF$ is the value of the essentially constant ind-object $F(L^\bullet)$ for these $s$ it follows that $H^i(RF(K^\bullet)) = 0$ for $i < 0$. \medskip\noindent To prove (2) we use the distinguished triangle $$ \tau_{\leq a}K^\bullet \to K^\bullet \to \tau_{\geq a + 1}K^\bullet \to (\tau_{\leq a}K^\bullet)[1] $$ of Remark \ref{remark-truncation-distinguished-triangle} to conclude via Lemma \ref{lemma-2-out-of-3-defined} that $RF$ is defined at $\tau_{\geq a + 1}K^\bullet$ as well and that we have a distinguished triangle $$ RF(\tau_{\leq a}K^\bullet) \to RF(K^\bullet) \to RF(\tau_{\geq a + 1}K^\bullet) \to RF(\tau_{\leq a}K^\bullet)[1] $$ in $D(\mathcal{B})$. By part (1) we see that $RF(\tau_{\geq a + 1}K^\bullet)$ has vanishing cohomology in degrees $< a + 1$. The long exact cohomology sequence of this distinguished triangle then shows what we want.
Let $A$ be an object of $\mathcal{A}$. Let $A[0] \to K^\bullet$ be any quasi-isomorphism. Then it is also true that $A[0] \to \tau_{\geq 0}K^\bullet$ is a quasi-isomorphism. Hence in the category $A[0]/\text{Qis}^{+}(\mathcal{A})$ the quasi-isomorphisms $s : A[0] \to K^\bullet$ with $K^n = 0$ for $n < 0$ are cofinal. Thus it is clear that $H^i(RF(A[0])) = 0$ for $i < 0$. Moreover, for such an $s$ the sequence $$ 0 \to A \to K^0 \to K^1 $$ is exact. Hence if $F$ is left exact, then $0 \to F(A) \to F(K^0) \to F(K^1)$ is exact as well, and we see that $F(A) \to H^0(F(K^\bullet))$ is an isomorphism for every $s : A[0] \to K^\bullet$ as above which implies that $H^0(RF(A[0])) = F(A)$. \medskip\noindent Let $0 \to A \to B \to C \to 0$ be a short exact sequence of $\mathcal{A}$. By Lemma \ref{lemma-derived-canonical-delta-functor} we obtain a distinguished triangle $(A[0], B[0], C[0], a, b, c)$ in $K^{+}(\mathcal{A})$. From the long exact cohomology sequence (and the vanishing for $i < 0$ proved above) we deduce that $0 \to R^0F(A) \to R^0F(B) \to R^0F(C)$ is exact. Hence $R^0F$ is left exact. Of course this also proves that if $F \to R^0F$ is an isomorphism, then $F$ is left exact.
If $A$ is right acyclic for $F$, then $RF(A[0]) = F(A)[0]$ and in particular $F(A) \to R^0F(A)$ is an isomorphism and $R^iF(A) = 0$ for $i \not = 0$. Conversely, if $F(A) \to R^0F(A)$ is an isomorphism and $R^iF(A) = 0$ for all $i > 0$ then $F(A[0]) \to RF(A[0])$ is a quasi-isomorphism by Lemma \ref{lemma-left-exact-higher-derived} part (1) and hence $A$ is acyclic. If $F$ is left exact then $F = R^0F$, see Lemma \ref{lemma-left-exact-higher-derived}.
By Lemma \ref{lemma-derived-canonical-delta-functor} we obtain a distinguished triangle $(A[0], B[0], C[0], a, b, c)$ in $K^{+}(\mathcal{A})$. As $RF$ is an exact functor and since $R^iF = 0$ for $i < 0$ and $R^0F = F$ (Lemma \ref{lemma-left-exact-higher-derived}) we obtain an exact cohomology sequence $$ 0 \to F(A) \to F(B) \to F(C) \to R^1F(A) \to \ldots $$ in the abelian category $\mathcal{B}$. Thus the lemma follows from the characterization of acyclic objects in Lemma \ref{lemma-F-acyclic}.
The functor $\mathcal{A} \to \text{Comp}^{+}(\mathcal{A})$, $A \mapsto A[0]$ is exact. The functor $\text{Comp}^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$ is a $\delta$-functor, see Lemma \ref{lemma-derived-canonical-delta-functor}. The functor $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is exact. Finally, the functor $H^0 : D^{+}(\mathcal{B}) \to \mathcal{B}$ is a homological functor, see Definition \ref{definition-unbounded-derived-category}. Hence we get the structure of a $\delta$-functor from Lemma \ref{lemma-compose-delta-functor-homological} and Lemma \ref{lemma-exact-compose-delta-functor}. Part (2) follows from Homology, Lemma \ref{homology-lemma-efface-implies-universal} and the description of acyclics in Lemma \ref{lemma-F-acyclic}.
First we claim the lemma holds for a bounded complex of acyclic objects. Namely, it holds for complexes with at most one nonzero object by definition. Suppose that $A^\bullet$ is a complex with $A^n = 0$ for $n \not \in [a, b]$. Using the ``stupid'' truncations we obtain a termwise split short exact sequence of complexes $$ 0 \to \sigma_{\geq a + 1} A^\bullet \to A^\bullet \to \sigma_{\leq a} A^\bullet \to 0 $$ see Homology, Section \ref{homology-section-truncations}. Thus a distinguished triangle $(\sigma_{\geq a + 1} A^\bullet, A^\bullet, \sigma_{\leq a} A^\bullet)$. By induction hypothesis the two outer complexes compute $RF$. Then the middle one does too by Lemma \ref{lemma-2-out-of-3-computes}. \medskip\noindent Suppose that $A^\bullet$ is a bounded below complex of acyclic objects. To show that $F(A) \to RF(A)$ is an isomorphism in $D^{+}(\mathcal{B})$ it suffices to show that $H^i(F(A)) \to H^i(RF(A))$ is an isomorphism for all $i$. Pick $i$. Consider the termwise split short exact sequence of complexes $$ 0 \to \sigma_{\geq i + 2} A^\bullet \to A^\bullet \to \sigma_{\leq i + 1} A^\bullet \to 0. $$ Note that this induces a termwise split short exact sequence $$ 0 \to \sigma_{\geq i + 2} F(A^\bullet) \to F(A^\bullet) \to \sigma_{\leq i + 1} F(A^\bullet) \to 0. $$ Hence we get distinguished triangles $$ \begin{matrix} (\sigma_{\geq i + 2} A^\bullet, A^\bullet, \sigma_{\leq i + 1} A^\bullet) \\ (\sigma_{\geq i + 2} F(A^\bullet), F(A^\bullet), \sigma_{\leq i + 1} F(A^\bullet)) \\ (RF(\sigma_{\geq i + 2} A^\bullet), RF(A^\bullet), RF(\sigma_{\leq i + 1} A^\bullet)) \end{matrix} $$ Using the last two we obtain a map of exact sequences $$ \xymatrix{ H^i(\sigma_{\geq i + 2} F(A^\bullet)) \ar[r] \ar[d] & H^i(F(A^\bullet)) \ar[r] \ar[d]^\alpha & H^i(\sigma_{\leq i + 1} F(A^\bullet)) \ar[r] \ar[d]^\beta & H^{i + 1}(\sigma_{\geq i + 2} F(A^\bullet)) \ar[d] \\ H^i(RF(\sigma_{\geq i + 2} A^\bullet)) \ar[r] & H^i(RF(A^\bullet)) \ar[r] & H^i(RF(\sigma_{\leq i + 1} A^\bullet)) \ar[r] & H^{i + 1}(RF(\sigma_{\geq i + 2} A^\bullet)) } $$ By the results of the first paragraph the map $\beta$ is an isomorphism. By inspection the objects on the upper left and the upper right are zero. Hence to finish the proof it suffices to show that $H^i(RF(\sigma_{\geq i + 2} A^\bullet)) = 0$ and $H^{i + 1}(RF(\sigma_{\geq i + 2} A^\bullet)) = 0$. This follows immediately from Lemma \ref{lemma-negative-vanishing}.
It is clear that $D_\mathcal{B}(\mathcal{A})$ is an additive subcategory preserved under the translation functors. If $X \oplus Y$ is in $D_\mathcal{B}(\mathcal{A})$, then both $H^n(X)$ and $H^n(Y)$ are kernels of maps between maps of objects of $\mathcal{B}$ as $H^n(X \oplus Y) = H^n(X) \oplus H^n(Y)$. Hence both $X$ and $Y$ are in $D_\mathcal{B}(\mathcal{A})$. By Lemma \ref{lemma-triangulated-subcategory} it therefore suffices to show that given a distinguished triangle $(X, Y, Z, f, g, h)$ such that $X$ and $Y$ are in $D_\mathcal{B}(\mathcal{A})$ then $Z$ is an object of $D_\mathcal{B}(\mathcal{A})$. The long exact cohomology sequence (\ref{equation-long-exact-cohomology-sequence-D}) and the definition of a weak Serre subcategory (see Homology, Definition \ref{homology-definition-serre-subcategory}) show that $H^n(Z)$ is an object of $\mathcal{B}$ for all $n$. Thus $Z$ is an object of $D_\mathcal{B}(\mathcal{A})$.
The functor $D(v) : D(\mathcal{A}) \to D(\mathcal{A}/\mathcal{B})$ is essentially surjective by Lemma \ref{lemma-derived-of-quotient}. For an object $X$ of $D(\mathcal{A})$ the adjunction mapping $c_X : uvX \to X$ maps to an isomorphism in $D(\mathcal{A}/\mathcal{B})$ because $vuv \cong v$ by the assumption that $vu \cong \text{id}$. Thus in a distinguished triangle $(uvX, X, Z, c_X, g, h)$ the object $Z$ is an object of $D_\mathcal{B}(\mathcal{A})$ as we see by looking at the long exact cohomology sequence. Hence $c_X$ is an element of the multiplicative system used to define the quotient category $D(\mathcal{A})/D_\mathcal{B}(\mathcal{A})$. Thus $uvX \cong X$ in $D(\mathcal{A})/D_\mathcal{B}(\mathcal{A})$. For $X, Y \in \Ob(\mathcal{A}))$ the map $$ \Hom_{D(\mathcal{A})/D_\mathcal{B}(\mathcal{A})}(X, Y) \longrightarrow \Hom_{D(\mathcal{A}/\mathcal{B})}(vX, vY) $$ is bijective because $u$ gives an inverse (by the remarks above).
Proof of (1). First choose an injection $A \to I^0$ of $A$ into an injective object of $\mathcal{A}$. Next, choose an injection $I_0/A \to I^1$ into an injective object of $\mathcal{A}$. Denote $d^0$ the induced map $I^0 \to I^1$. Next, choose an injection $I^1/\Im(d^0) \to I^2$ into an injective object of $\mathcal{A}$. Denote $d^1$ the induced map $I^1 \to I^2$. And so on. By Lemma \ref{lemma-cohomology-bounded-below} part (2) follows from part (3). Part (3) is a special case of Lemma \ref{lemma-subcategory-right-resolution}.
The ``correct'' proof of part (1) is explained in Remark \ref{remark-easier-proofs}. We also give a direct proof here. \medskip\noindent We first show that (2) implies (1). Namely, let $\tilde \alpha : K \to \tilde L^\bullet$, $\pi$, $s$ be as in Lemma \ref{lemma-make-injective}. Since $\tilde \alpha$ is injective by (2) there exists a morphism $\tilde \beta : \tilde L^\bullet \to I^\bullet$ such that $\gamma = \tilde \beta \circ \tilde \alpha$. Set $\beta = \tilde \beta \circ s$. Then we have $$ \beta \circ \alpha = \tilde \beta \circ s \circ \pi \circ \tilde \alpha \sim \tilde \beta \circ \tilde \alpha = \gamma $$ as desired. \medskip\noindent Assume that $\alpha : K^\bullet \to L^\bullet$ is injective. Suppose we have already defined $\beta$ in all degrees $\leq n - 1$ compatible with differentials and such that $\gamma^j = \beta^j \circ \alpha^j$ for all $j \leq n - 1$. Consider the commutative solid diagram $$ \xymatrix{ K^{n - 1} \ar[r] \ar@/_2pc/[dd]_\gamma \ar[d]^\alpha & K^n \ar@/^2pc/[dd]^\gamma \ar[d]^\alpha \\ L^{n - 1} \ar[r] \ar[d]^\beta & L^n \ar@{-->}[d] \\ I^{n - 1} \ar[r] & I^n } $$ Thus we see that the dotted arrow is prescribed on the subobjects $\alpha(K^n)$ and $d^{n - 1}(L^{n - 1})$. Moreover, these two arrows agree on $\alpha(d^{n - 1}(K^{n - 1}))$. Hence if \begin{equation} \label{equation-qis} \alpha(d^{n - 1}(K^{n - 1})) = \alpha(K^n) \cap d^{n - 1}(L^{n - 1}) \end{equation} then these morphisms glue to a morphism $\alpha(K^n) + d^{n - 1}(L^{n - 1}) \to I^n$ and, using the injectivity of $I^n$, we can extend this to a morphism from all of $L^n$ into $I^n$. After this by induction we get the morphism $\beta$ for all $n$ simultaneously (note that we can set $\beta^n = 0$ for all $n \ll 0$ since $I^\bullet$ is bounded below -- in this way starting the induction). \medskip\noindent It remains to prove the equality (\ref{equation-qis}). The reader is encouraged to argue this for themselves with a suitable diagram chase. Nonetheless here is our argument. Note that the inclusion $\alpha(d^{n - 1}(K^{n - 1})) \subset \alpha(K^n) \cap d^{n - 1}(L^{n - 1})$ is obvious. Take an object $T$ of $\mathcal{A}$ and a morphism $x : T \to L^n$ whose image is contained in the subobject $\alpha(K^n) \cap d^{n - 1}(L^{n - 1})$. Since $\alpha$ is injective we see that $x = \alpha \circ x'$ for some $x' : T \to K^n$. Moreover, since $x$ lies in $d^{n - 1}(L^{n - 1})$ we see that $d^n \circ x = 0$. Hence using injectivity of $\alpha$ again we see that $d^n \circ x' = 0$. Thus $x'$ gives a morphism $[x'] : T \to H^n(K^\bullet)$. On the other hand the corresponding map $[x] : T \to H^n(L^\bullet)$ induced by $x$ is zero by assumption. Since $\alpha$ is a quasi-isomorphism we conclude that $[x'] = 0$. This of course means exactly that the image of $x'$ is contained in $d^{n - 1}(K^{n - 1})$ and we win.
This follows from Remark \ref{remark-easier-proofs}. We also give a direct argument here. \medskip\noindent Let $\tilde \alpha : K \to \tilde L^\bullet$, $\pi$, $s$ be as in Lemma \ref{lemma-make-injective}. If we can show that $\beta_1 \circ\pi$ is homotopic to $\beta_2 \circ \pi$, then we deduce that $\beta_1 \sim \beta_2$ because $\pi \circ s$ is the identity. Hence we may assume $\alpha^n : K^n \to L^n$ is the inclusion of a direct summand for all $n$. Thus we get a short exact sequence of complexes $$ 0 \to K^\bullet \to L^\bullet \to M^\bullet \to 0 $$ which is termwise split and such that $M^\bullet$ is acyclic. We choose splittings $L^n = K^n \oplus M^n$, so we have $\beta_i^n : K^n \oplus M^n \to I^n$ and $\gamma^n : K^n \to I^n$. In this case the condition on $\beta_i$ is that there are morphisms $h_i^n : K^n \to I^{n - 1}$ such that $$ \gamma^n - \beta_i^n|_{K^n} = d \circ h_i^n + h_i^{n + 1} \circ d $$ Thus we see that $$ \beta_1^n|_{K^n} - \beta_2^n|_{K^n} = d \circ (h_1^n - h_2^n) + (h_1^{n + 1} - h_2^{n + 1}) \circ d $$ Consider the map $h^n : K^n \oplus M^n \to I^{n - 1}$ which equals $h_1^n - h_2^n$ on the first summand and zero on the second. Then we see that $$ \beta_1^n - \beta_2^n - (d \circ h^n + h^{n + 1}) \circ d $$ is a morphism of complexes $L^\bullet \to I^\bullet$ which is identically zero on the subcomplex $K^\bullet$. Hence it factors as $L^\bullet \to M^\bullet \to I^\bullet$. Thus the result of the lemma follows from Lemma \ref{lemma-acyclic-is-zero}.
Let $a$ be an element of the right hand side. We may represent $a = \gamma\alpha^{-1}$ where $\alpha : K^\bullet \to L^\bullet$ is a quasi-isomorphism and $\gamma : K^\bullet \to I^\bullet$ is a map of complexes. By Lemma \ref{lemma-morphisms-lift} we can find a morphism $\beta : L^\bullet \to I^\bullet$ such that $\beta \circ \alpha$ is homotopic to $\gamma$. This proves that the map is surjective. Let $b$ be an element of the left hand side which maps to zero in the right hand side. Then $b$ is the homotopy class of a morphism $\beta : L^\bullet \to I^\bullet$ such that there exists a quasi-isomorphism $\alpha : K^\bullet \to L^\bullet$ with $\beta  \circ \alpha$ homotopic to zero. Then Lemma \ref{lemma-morphisms-equal-up-to-homotopy} shows that $\beta$ is homotopic to zero also, i.e., $b = 0$.
Step 1. Choose an injective resolution $A^\bullet \to I^\bullet$ (see Lemma \ref{lemma-injective-resolutions-exist}) or use the given one. Recall that $\text{Comp}^{+}(\mathcal{A})$ is an abelian category, see Homology, Lemma \ref{homology-lemma-cat-cochain-abelian}. Hence we may form the pushout along the injective map $A^\bullet \to I^\bullet$ to get $$ \xymatrix{ 0 \ar[r] & A^\bullet \ar[r] \ar[d] & B^\bullet \ar[r] \ar[d] & C^\bullet \ar[r] \ar[d] & 0 \\ 0 \ar[r] & I^\bullet \ar[r] & E^\bullet \ar[r] & C^\bullet \ar[r] & 0 } $$ Note that the lower short exact sequence is termwise split, see Homology, Lemma \ref{homology-lemma-characterize-injectives}. Hence it suffices to prove the lemma when $0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$ is termwise split. \medskip\noindent Step 2. Choose splittings. In other words, write $B^n = A^n \oplus C^n$. Denote $\delta : C^\bullet \to A^\bullet[1]$ the morphism as in Homology, Lemma \ref{homology-lemma-ses-termwise-split-cochain}. Choose injective resolutions $f_1 : A^\bullet \to I_1^\bullet$ and $f_3 : C^\bullet \to I_3^\bullet$. (If $A^\bullet$ is a complex of injectives, then use $I_1^\bullet = A^\bullet$.) We may assume $f_3$ is injective in every degree. By Lemma \ref{lemma-morphisms-lift} we may find a morphism $\delta' : I_3^\bullet \to I_1^\bullet[1]$ such that $\delta' \circ f_3 = f_1[1] \circ \delta$ (equality of morphisms of complexes). Set $I_2^n = I_1^n \oplus I_3^n$. Define $$ d_{I_2}^n = \left( \begin{matrix} d_{I_1}^n & (\delta')^n \\ 0 & d_{I_3}^n \end{matrix} \right) $$ and define the maps $B^n \to I_2^n$ to be given as the sum of the maps $A^n \to I_1^n$ and $C^n \to I_3^n$. Everything is clear.
The first equality follows from Lemma \ref{lemma-morphisms-from-projective-complex}. Note that there is a distinguished triangle $$ (\tau_{\leq n - 1}K^\bullet, K^\bullet, \tau_{\geq n}K^\bullet, f, g, h) $$ by Remark \ref{remark-truncation-distinguished-triangle}. Hence, by Lemma \ref{lemma-representable-homological} it suffices to prove $\Hom_{K(\mathcal{A})}(P^\bullet, \tau_{\leq n - 1}K^\bullet) = 0$ and $\Hom_{K(\mathcal{A})}(P^\bullet, \tau_{\geq n} K^\bullet) = 0$. The first vanishing is trivial and the second is Lemma \ref{lemma-projective-into-acyclic-is-zero}.
Part (2) is a direct consequences of part (1) and Definition \ref{definition-derived-functor}. To prove (1) let $\alpha : I^\bullet \to K^\bullet$ be a quasi-isomorphism into a complex. By Lemma \ref{lemma-morphisms-lift} we see that $\alpha$ has a left inverse. Hence the category $I^\bullet/\text{Qis}^{+}(\mathcal{A})$ is essentially constant with value $\text{id} : I^\bullet \to I^\bullet$. Thus also the ind-object $$ I^\bullet/\text{Qis}^{+}(\mathcal{A}) \longrightarrow \mathcal{D}, \quad (I^\bullet \to K^\bullet) \longmapsto F(K^\bullet) $$ is essentially constant with value $F(I^\bullet)$. This proves (1), see Definitions \ref{definition-right-derived-functor-defined} and \ref{definition-computes}.
Combine Lemma \ref{lemma-injective-acyclic} and Proposition \ref{proposition-enough-acyclics} for the second assertion. To see the first assertion combine Lemma \ref{lemma-injective-resolutions-exist}, Lemma \ref{lemma-injective-acyclic}, Lemma \ref{lemma-existence-computes}, and Equation (\ref{equation-everywhere}).
This lemma simply reviews some of the results obtained so far. Note that by Lemma \ref{lemma-enough-injectives-right-derived} $RF$ is everywhere defined. Here are some references: \begin{enumerate} \item The derived functor is exact: This boils down to Lemma \ref{lemma-2-out-of-3-defined}. \item This is true because $K^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$ is exact and compositions of exact functors are exact. \item This is true because $\text{Comp}^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$ is a $\delta$-functor, see Lemma \ref{lemma-derived-canonical-delta-functor}. \item This is true because $\mathcal{A} \to \text{Comp}^{+}(\mathcal{A})$ is exact and precomposing a $\delta$-functor by an exact functor gives a $\delta$-functor. \end{enumerate}
This lemma simply reviews some of the results obtained so far. Note that by Lemma \ref{lemma-enough-injectives-right-derived} $RF$ is everywhere defined. Here are some references: \begin{enumerate} \item This follows from Lemma \ref{lemma-right-derived-properties} part (3) combined with the long exact cohomology sequence (\ref{equation-long-exact-cohomology-sequence-D}) for $D^{+}(\mathcal{B})$. \item This is Lemma \ref{lemma-left-exact-higher-derived}. \item This is the fact that injective objects are acyclic. \item This is Lemma \ref{lemma-right-derived-delta-functor}. \end{enumerate}