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In what follows I am going to assume $\mathcal{T}$ is the homotopy category of a stably symmetric monoidal ∞-category. The argument I am going to give can probably be generalized to any case of intere …

I'm going to do a proof assuming we are in a stable $\infty$-category (I'm pretty sure this is almost equivalent to your "sufficiently rich" situation anyway). In your case $F=j_!j^!$ and $G=i_*i^*$. …

Since I have already given a similar answer recently, I don't want to be branded as the "anti-triangular" guy: the formalism of triangulated categories can be useful in certain settings. That said the …

No. Let $j:\mathbb{A}^2_k\smallsetminus\{0\}\to \mathbb{A}^2_k$ be the canonical open embedding. Then the derived pushforward $Rj_*$ is fully faithful and colimit-preserving. In particular, the subc …

My opinion, and that of many other people although not of everyone, is that the "correct" notion is that of stable ∞-category. Now, this is not a category in the strictest sense, rather a generalizat …

I do not know of an answer for a general triangulated category (non-topological triangulated categories are very unusual), but as soon as you ask for some more structure the thesis follows very quickl …

The key here is that $SH^{eff}(S)$ is closed under suspensions, so there's an inclusion $j_{n+1}:\Sigma^{n+1}_T SH^{eff}(S)\subseteq \Sigma^n_T SH^{eff}(S)$. Hence you can write $i_{n+1}=i_n \circ j_{ …