**Proof.**
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 $H^0(RF(A[0])) = F(A)$.

Let $0 \to A \to B \to C \to 0$ be a short exact sequence of $\mathcal{A}$. By Lemma 13.12.1 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.
$\square$

## Comments (1)

Comment #7814 by Anonymous on