The Stacks project

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 $K^+(\mathcal{A})$".

Thanks and fixed here.

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.

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