Lemma 13.16.5. Let $F : \mathcal{A} \to \mathcal{B}$ be a left exact functor between abelian categories and assume $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is everywhere defined. Let $0 \to A \to B \to C \to 0$ be a short exact sequence of $\mathcal{A}$.

1. If $A$ and $C$ are right acyclic for $F$ then so is $B$.

2. If $A$ and $B$ are right acyclic for $F$ then so is $C$.

3. If $B$ and $C$ are right acyclic for $F$ and $F(B) \to F(C)$ is surjective then $A$ is right acyclic for $F$.

In each of the three cases

$0 \to F(A) \to F(B) \to F(C) \to 0$

is a short exact sequence of $\mathcal{B}$.

Proof. By Lemma 13.12.1 we obtain a distinguished triangle $(A, B, C, a, b, c)$ in $K^{+}(\mathcal{A})$. As $RF$ is an exact functor and since $R^ iF = 0$ for $i < 0$ and $R^0F = F$ (Lemma 13.16.3) we obtain an exact cohomology sequence

$0 \to F(A) \to F(B) \to F(C) \to R^1F(A) \to \ldots$

in the abelian category $\mathcal{B}$. Thus the lemma follows from the characterization of acyclic objects in Lemma 13.16.4. $\square$

Comment #8401 by on

In the first sentence, I believe it should be $D^+(\mathcal{A})$, not $K^+(\mathcal{A})$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).