Lemma 13.16.3 (05TD)—The Stacks project

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.

We have $R^ iF = 0$ for $i < 0$,
$R^0F$ is left exact,
the map $F \to R^0F$ is an isomorphism if and only if $F$ is left exact.

Proof. Let $A$ be an object of $\mathcal{A}$. By Lemma 13.16.1 we have $H^ i(RF(A[0]) = 0$ for $i < 0$. This proves (1).

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 $D^{+}(\mathcal{A})$. 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.

Assume $F$ is left exact. Recall that $RF(A[0])$ is the value of the essentially constant system $F(K^\bullet )$ for $s : A[0] \to K^\bullet $ quasi-isomorphisms. It follows that $R^0F(A)$ is the value of the essentially constant system $H^0(F(K^\bullet ))$ for $s : A[0] \to K^\bullet $ quasi-isomorphisms, see Categories, Lemma 4.22.8. But if $s : A[0] \to K^\bullet $ is a quasi-isomorphism, then $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. It follows from Categories, Lemma 4.22.11 that we may restrict to such $s$. Moreover, for such an $s$ the sequence

\[ 0 \to A \to K^0 \to K^1 \]

is exact. Since $F$ is left exact we see that $0 \to F(A) \to F(K^0) \to F(K^1)$ is exact as well. It follows that $F(A) \to H^0(F(K^\bullet ))$ is an isomorphism and the system is actually constant with value $F(A)$. We conclude $R^0F = F$ as desired. $\square$

Comments (4)

Comment #7814 by Anonymous on October 03, 2022 at 11:09

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 #8041 by Stacks Project on January 19, 2023 at 09:19

Thanks and fixed here.

Comment #8400 by Elías Guisado on May 16, 2023 at 12:20

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 $A[0]\to K^\bullet$ , with $K^n=0$ for $n<0$ and a map $RF(A[0])\to F(K^\bullet)$ satisfying Categories, Definition 4.22.1, (1). In particular, the composite $RF(A[0])\to F(K^\bullet)\to RF(A[0])$ is the identity and there are $K^\bullet\to L^\bullet$ and $A[0]\to L^\bullet$ in $A[0]/\text{Qis}^+(\mathcal{A})$ , with $L^n=0$ for $n<0$ , such that, inside $\mathcal{D}^+(\mathcal{B})$ , the map $F(A[0])\to F(L^\bullet)$ equals the composite $F(A[0])\to RF(A[0])\to F(K^\bullet)\to F(L^\bullet)$ . 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 $F(A[0])\to F(L^\bullet)$ and $F(K^\bullet)\to F(L^\bullet)$ are isos. This implies that the image under $H^0$ of $RF(A[0])\to F(K^\bullet)$ must be an epimorphism. But it is also a split monomorphism; hence, it is an iso, and we win.

Comment #9012 by Stacks project on April 17, 2024 at 15:45

Thanks! I fixed it in a slightly different way because it suffered from the same pitfall as the previous lemma. See changes.

Comments (4)

Post a comment