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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