By part (2) of Definition \ref{definition-siblings} the image of the functor $F'$ is contained in the essential image of the functor $F$. Hence the functor $H = F^{-1} \circ F'$ is a sibling of the identity functor. This reduces us to the case described in the next paragraph. \medskip\noindent Let $\mathcal{D} = D^b(\mathcal{A})$. We have to show a sibling $F : \mathcal{D} \to \mathcal{D}$ of the identity functor is isomorphic to the identity functor. Given an object $X$ of $\mathcal{D}$ let us say $X$ has {\it width} $w = w(X)$ if $w \geq 0$ is minimal such that there exists an integer $a \in \mathbf{Z}$ with $H^i(X) = 0$ for $i \not \in [a, a + w - 1]$. Since $F$ is a sibling of the identity and since $F \circ [n] = [n] \circ F$ we are aready given isomorphisms $$ c_X : X \to F(X) $$ for $w(X) \leq 1$ compatible with shifts. Moreover, if $X = A[-a]$ and $X' = A'[-a]$ for some $A, A' \in \Ob(\mathcal{A})$ then for any morphism $f : X \to X'$ the diagram \begin{equation} \label{equation-to-show} \vcenter{ \xymatrix{ X \ar[d]_{c_X} \ar[r]_f & X' \ar[d]^{c_{X'}} \\ F(X) \ar[r]^{F(f)} & F(X') } } \end{equation} is commutative. \medskip\noindent Next, let us show that for any morphism $f : X \to X'$  with $w(X), w(X') \leq 1$ the diagram (\ref{equation-to-show}) commutes. If $X$ or $X'$ is zero, this is clear. If not then we can write $X = A[-a]$ and $X' = A'[-a']$ for unique $A, A'$ in $\mathcal{A}$ and $a, a' \in \mathbf{Z}$. The case $a = a'$ was discussed above. If $a' > a$, then $f = 0$ (Derived Categories, Lemma \ref{derived-lemma-negative-exts}) and the result is clear. If $a' < a$ then $f$ corresponds to an element $\xi \in \Ext^q(A, A')$ with $q = a - a'$. Using Yoneda extensions, see Derived Categories, Section \ref{derived-section-ext}, we can find $A = A_0, A_1, \ldots, A_{q - 1}, A_q = A' \in \Ob(\mathcal{A})$ and elements $$ \xi_i \in \Ext^1(A_{i - 1}, A_i) $$ such that $\xi$ is the composition $\xi_q \circ \ldots \circ \xi_1$. In other words, setting $X_i = A_i[-a + i]$ we obtain morphisms $$ X = X_0 \xrightarrow{f_1} X_1 \to \ldots \to X_{q - 1} \xrightarrow{f_q} X_q = X' $$ whose compostion is $f$. Since the commutativity of  (\ref{equation-to-show}) for $f_1, \ldots, f_q$ implies it for $f$, this reduces us to the case $q = 1$. In this case after shifting we may assume we have a distinguished triangle $$ A' \to E \to A \xrightarrow{f} A'[1] $$ Observe that $E$ is an object of $\mathcal{A}$. Consider the following diagram $$ \xymatrix{ E \ar[d]_{c_E} \ar[r] & A \ar[d]_{c_A} \ar[r]_f & A'[1] \ar[d]^{c_{A'}[1]} \ar@{..>}@<-1ex>[d]_\gamma \ar@{..>}[ld]^\epsilon \ar[r] & E[1] \ar[d]^{c_E[1]} \\ F(E) \ar[r] & F(A) \ar[r]^{F(f)} & F(A')[1] \ar[r] & F(E)[1] } $$ whose rows are distinguished triangles. The square on the right commutes already but we don't yet know that the middle square does. By the axioms of a triangulated category we can find a morphism $\delta$ which does make the diagram commute. Then $\gamma - c_{A'}[1]$ composed with $F(A')[1] \to F(E)[1]$ is zero hence we can find $\epsilon : A'[1] \to F(A)$ such that $\gamma - c_{A'}[1] = F(f) \circ \epsilon$. However, any arrow $A'[1] \to F(A)$ is zero as it is a negative ext class between objects of $\mathcal{A}$. Hence $\gamma = c_{A'}[1]$ and we conclude the middle square commutes too which is what we wanted to show. \medskip\noindent To finish the proof we are going to argue by induction on $w$ that there exist isomorphisms $c_X : X \to F(X)$ for all $X$ with $w(X) \leq w$ compatible with all morphisms between such objects. The base case $w = 1$ was shown above. Assume we know the result for some $w \geq 1$. \medskip\noindent Let $X$ be an object with $w(X) = w + 1$. Pick $a \in \mathbf{Z}$ with $H^i(X) = 0$ for $i \not \in [a, a + w]$. Set $b = a + w$ so that $H^b(X)$ is nonzero. Pick $N$ in $\mathcal{A}$ such that there exists a surjection $N \to H^b(X)$, such that $\Hom(H^b(X), N) = 0$ and such that $\Ext^q(N, H^i(X)) = 0$ for $i \in \mathbf{Z}$ and $q > 0$. This is possible because $\mathcal{A}$ has enough negative objects by appplying the definition to $\bigoplus H^i(X)$. By the vanishing of Exts we can lift the surjection $N \to H^b(X)$ to a morphism $N[-b] \to X$; details omitted. Let us call a morphism $N[-b] \to X$ constructed in this manner a {\it good morphism}. Given a good morphism $N[-b] \to X$ choose a distinguished diagram $$ N[-b] \to X \to Y \to N[-b + 1] $$ Computing the long exact cohomology sequence we find $w(Y) \leq w$. Hence by induction we find the solid arrows in the following diagram $$ \xymatrix{ N[-b] \ar[r] \ar[d]_{c_N[-b]} & X \ar[r] \ar@{..>}[d]_{c_{N[-b] \to X}} & Y \ar[r] \ar[d]^{c_Y} & N[-b + 1] \ar[d]^{c_N[-b + 1]} \\ F(N)[-b] \ar[r] & F(X) \ar[r] & F(Y) \ar[r] & F(N)[-b + 1] } $$ We obtain the dotted arrow $c_{N[-b] \to X}$. By Derived Categories, Lemma \ref{derived-lemma-uniqueness-third-arrow} the dotted arrow is unique because $\Hom(X, F(N)[-b]) \cong \Hom(X, N[-b]) = 0$ by our choice of $N$. In fact, $c_{N[-b] \to X}$ is the unique dotted arrow making the square with vertices $X, Y, F(X), F(Y)$ commute. Our goal is to show that $c_{N[-b] \to X}$ is independent of the choice of good morphism $N[-b] \to X$ and that the diagrams (\ref{equation-to-show}) commute. \medskip\noindent Independence of the choice of good morphism. Given two good morphisms $N[-b] \to X$ and $N'[-b] \to X$ we get another good morphism, namely $(N \oplus N')[-b] \to X$. Thus we may assume $N'[-b] \to X$ factors as $N'[-b] \to N[-b] \to X$ for some morphism $N' \to N$. Choose distinguished triangles $N[-b] \to X \to Y \to N[-b + 1]$ and $N'[-b] \to X \to Y' \to N'[-b + 1]$. By axiom TR3 we can find a morphism $g : Y' \to Y$ which joint with $\text{id}_X$ and $N' \to N$ forms a morphism of triangles. Since we have (\ref{equation-to-show}) for $g$ we conclude that $$ (F(X) \to F(Y)) \circ c_{N'[-b] \to X} = (F(X) \to F(Y)) \circ c_{N[-b] \to X} $$ The uniqueness of $c_{N[-b] \to X}$ pointed out in the construction above now shows that $c_{N'[-b] \to X} = c_{N[-b] \to X}$. \medskip\noindent Let $f : X \to X'$ be a morphism of objects with $w(X) \leq w + 1$ and $w(X') \leq w + 1$. Choose $a \leq b \leq a + w$ such that $H^i(X) = 0$ for $i \not \in [a, b]$ and $a' \leq b' \leq a' + w$ such that $H^i(X') = 0$ for $i \not \in [a', b']$. We will use induction on $(b' - a') + (b - a)$ to show this. (The base case is when this number is zero which is OK because $w \geq 1$.) We distinguish two cases. \medskip\noindent Case I: $b' < b$. In this case we choose a good morphism $N[-b] \to X$ such that in addition $\Ext^q(N, H^i(X')) = 0$ for $q > 0$ and all $i$. Choose a distuiguished triangle $N[-b] \to X \to Y \to N[-b + 1]$. Since $\Hom(N[-b], X') = 0$ by our choice of $N$ and we find that $f$ factors as $X \to Y \to X'$. Since $H^i(Y)$ is nonzero only for $i \in [a, b - 1]$ we see by induction that (\ref{equation-to-show}) commutes for $Y \to X'$. The diagram (\ref{equation-to-show}) commutes for $X \to Y$ by construction if $w(X) = w + 1$ and by our first induction hypothesis if $w(X) \leq w$. Hence (\ref{equation-to-show}) commutes for $f$. \medskip\noindent Case II: $b' \geq b$. In this case we choose a good morphism $N'[-b'] \to X'$ such that $\Hom(H^{b'}(X), N') = 0$ (this is relevant only if $b' = b$). We choose a distinguished triangle $N'[-b'] \to X' \to Y' \to N'[-b' + 1]$. Since $\Hom(X, X') \to \Hom(X, Y')$ is injective by our choice of $N'$ (details omitted) the same is true for $\Hom(X, F(X')) \to \Hom(X, F(Y'))$. Hence it suffices in this case to check that (\ref{equation-to-show}) commutes for the composition $X \to Y'$ of the morphisms $X \to X' \to Y'$. Since $H^i(Y')$ is nonzero only for $i \in [a', b' - 1]$ we conclude by induction hypothesis.