Lemma 13.16.3. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor between abelian categories and assume $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is everywhere defined.

1. We have $R^ iF = 0$ for $i < 0$,

2. $R^0F$ is left exact,

3. the map $F \to R^0F$ is an isomorphism if and only if $F$ is left exact.

Proof. Let $A$ be an object of $\mathcal{A}$. Let $A \to K^\bullet$ be any quasi-isomorphism. Then it is also true that $A \to \tau _{\geq 0}K^\bullet$ is a quasi-isomorphism. Hence in the category $A/\text{Qis}^{+}(\mathcal{A})$ the quasi-isomorphisms $s : A \to K^\bullet$ with $K^ n = 0$ for $n < 0$ are cofinal. Thus it is clear that $H^ i(RF(A)) = 0$ 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 $s : A \to K^\bullet$ as above which implies that $H^0(RF(A)) = F(A)$.

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 $(A, B, C, a, b, c)$ in $D^{+}(\mathcal{A})$. 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. $\square$

Comment #7814 by Anonymous on

In the second sentence of the second paragraph of the proof, I think it should read "... we obtain a distinguished triangle ... in $D^+(\mathcal{A})$" instead of "... we obtain a distinguished triangle ... in $K^+(\mathcal{A})$".

Comment #8400 by on

I don't know if it is worth to add any of this to the already existing proof, but in case it helps anyone (and maybe even my future self), here are additonal details to conclude that $F\to R^0F$ is an isomorphism provided that $F$ is left-exact: by what is already written (and maybe 17.5), there is a quasi-isomorphism $A\to K^\bullet$, with $K^n=0$ for $n<0$ and a map $RF(A)\to F(K^\bullet)$ satisfying Categories, Definition 4.22.1, (1). In particular, the composite $RF(A)\to F(K^\bullet)\to RF(A)$ is the identity and there are $K^\bullet\to L^\bullet$ and $A\to L^\bullet$ in $A/\text{Qis}^+(\mathcal{A})$, with $L^n=0$ for $n<0$, such that, inside $\mathcal{D}^+(\mathcal{B})$, the map $F(A)\to F(L^\bullet)$ equals the composite $F(A)\to RF(A)\to F(K^\bullet)\to F(L^\bullet)$. We claim that the image under $H^0$ of each of the maps in the last composite is an isomorphism (the first one gives $F(A)\to R^0F(A)$). Note that by what is already written in the proof, the images under $H^0$ of $F(A)\to F(L^\bullet)$ and $F(K^\bullet)\to F(L^\bullet)$ are isos. This implies that the image under $H^0$ of $RF(A)\to F(K^\bullet)$ must be an epimorphism. But it is also a split monomorphism; hence, it is an iso, and we win.

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