**Proof.**
If $RF$ is defined at $X$ and $Y$, then the distinguished triangle $X \to X \oplus Y \to Y \to X[1]$ (Lemma 13.4.11) and Lemma 13.14.6 shows that $RF$ is defined at $X \oplus Y$ and that we have a distinguished triangle $RF(X) \to RF(X \oplus Y) \to RF(Y) \to RF(X)[1]$. Applying Lemma 13.4.11 to this once more we find that $RF(X \oplus Y) = RF(X) \oplus RF(Y)$. This proves (1) and the final assertion.

Conversely, assume that $RF$ is defined at $X \oplus Y$ and that $\mathcal{D}'$ is Karoubian. Since $S$ is a saturated system $S$ is the set of arrows which become invertible under the additive localization functor $Q : \mathcal{D} \to S^{-1}\mathcal{D}$, see Categories, Lemma 4.27.21. Thus for any $s : X \to X'$ and $s' : Y \to Y'$ in $S$ the morphism $s \oplus s' : X \oplus Y \to X' \oplus Y'$ is an element of $S$. In this way we obtain a functor

\[ X/S \times Y/S \longrightarrow (X \oplus Y)/S \]

Recall that the categories $X/S, Y/S, (X \oplus Y)/S$ are filtered (Categories, Remark 4.27.7). By Categories, Lemma 4.22.12 $X/S \times Y/S$ is filtered and $F|_{X/S} : X/S \to \mathcal{D}'$ (resp. $G|_{Y/S} : Y/S \to \mathcal{D}'$) is essentially constant if and only if $F|_{X/S} \circ \text{pr}_1 : X/S \times Y/S \to \mathcal{D}'$ (resp. $G|_{Y/S} \circ \text{pr}_2 : X/S \times Y/S \to \mathcal{D}'$) is essentially constant. Below we will show that the displayed functor is cofinal, hence by Categories, Lemma 4.22.11. we see that $F|_{(X \oplus Y)/S}$ is essentially constant implies that $F|_{X/S} \circ \text{pr}_1 \oplus F|_{Y/S} \circ \text{pr}_2 : X/S \times Y/S \to \mathcal{D}'$ is essentially constant. By Homology, Lemma 12.30.3 (and this is where we use that $\mathcal{D}'$ is Karoubian) we see that $F|_{X/S} \circ \text{pr}_1 \oplus F|_{Y/S} \circ \text{pr}_2$ being essentially constant implies $F|_{X/S} \circ \text{pr}_1$ and $F|_{Y/S} \circ \text{pr}_2$ are essentially constant proving that $RF$ is defined at $X$ and $Y$.

Proof that the displayed functor is cofinal. To do this pick any $t : X \oplus Y \to Z$ in $S$. Using MS2 we can find morphisms $Z \to X'$, $Z \to Y'$ and $s : X \to X'$, $s' : Y \to Y'$ in $S$ such that

\[ \xymatrix{ X \ar[d]^ s & X \oplus Y \ar[d] \ar[l] \ar[r] & Y \ar[d]_{s'} \\ X' & Z \ar[l] \ar[r] & Y' } \]

commutes. This proves there is a map $Z \to X' \oplus Y'$ in $(X \oplus Y)/S$, i.e., we get part (1) of Categories, Definition 4.17.1. To prove part (2) it suffices to prove that given $t : X \oplus Y \to Z$ and morphisms $s_ i \oplus s'_ i : Z \to X'_ i \oplus Y'_ i$, $i = 1, 2$ in $(X \oplus Y)/S$ we can find morphisms $a : X'_1 \to X'$, $b : X'_2 \to X'$, $c : Y'_1 \to Y'$, $d : Y'_2 \to Y'$ in $S$ such that $a \circ s_1 = b \circ s_2$ and $c \circ s'_1 = d \circ s'_2$. To do this we first choose any $X'$ and $Y'$ and maps $a, b, c, d$ in $S$; this is possible as $X/S$ and $Y/S$ are filtered. Then the two maps $a \circ s_1, b \circ s_2 : Z \to X'$ become equal in $S^{-1}\mathcal{D}$. Hence we can find a morphism $X' \to X''$ in $S$ equalizing them. Similarly we find $Y' \to Y''$ in $S$ equalizing $c \circ s'_1$ and $d \circ s'_2$. Replacing $X'$ by $X''$ and $Y'$ by $Y''$ we get $a \circ s_1 = b \circ s_2$ and $c \circ s'_1 = d \circ s'_2$.

The proof of the corresponding statements for $LF$ are dual.
$\square$

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