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Let $\mathcal{A}$ be an abelian category and consider the following categories:

  • $\mathbf{Ch} (\mathcal{A})$, the category of cochain complexes in $\mathcal{A}$.
  • The full subcategories $\mathbf{Ch}^\mathrm{b} (\mathcal{A})$ (cochains bounded), $\mathbf{Ch}^- (\mathcal{A})$ (cochains bounded above), and $\mathbf{Ch}^+ (\mathcal{A})$ (cochains bounded below).
  • The derived category $\mathbf{D} (\mathcal{A})$, i.e. the category of cochain complexes in $\mathcal{A}$ localised with respect to quasi-isomorphisms.
  • The full subcategories $\mathbf{D}^\mathrm{b} (\mathcal{A})$ (cohomology bounded), $\mathbf{D}^- (\mathcal{A})$ (cohomology bounded above), and $\mathbf{D}^+ (\mathcal{A})$ (cohomology bounded below).
  • $\mathbf{Q}^{-1} \mathbf{Ch}^\mathrm{b} (\mathcal{A})$, the category of bounded cochain complexes localised with respect to quasi-isomorphisms, and analogously $\mathbf{Q}^{-1} \mathbf{Ch}^- (\mathcal{A})$ and $\mathbf{Q}^{-1} \mathbf{Ch}^+ (\mathcal{A})$.

By the universal property of localisation, the inclusion $\mathbf{Ch}^\mathrm{b} (\mathcal{A}) \hookrightarrow \mathbf{Ch} (\mathcal{A})$ induces a functor $\mathbf{Q}^{-1} \mathbf{Ch}^\mathrm{b} (\mathcal{A}) \to \mathbf{D} (\mathcal{A})$; similarly we have functors $\mathbf{Q}^{-1} \mathbf{Ch}^- (\mathcal{A}) \to \mathbf{D} (\mathcal{A})$ and $\mathbf{Q}^{-1} \mathbf{Ch}^+ (\mathcal{A}) \to \mathbf{D} (\mathcal{A})$. It is easy to see that a cochain complex whose cohomology is bounded (resp. bounded above, bounded below) is quasi-isomorphic to a cochain complex that is itself bounded (resp. bounded above, bounded below). Thus, these functors factor as essentially surjective functors $\mathbf{Q}^{-1} \mathbf{Ch}^\mathrm{b} (\mathcal{A}) \to \mathbf{D}^\mathrm{b} (\mathcal{A})$, $\mathbf{Q}^{-1} \mathbf{Ch}^- (\mathcal{A}) \to \mathbf{D}^- (\mathcal{A})$ and $\mathbf{Q}^{-1} \mathbf{Ch}^+ (\mathcal{A}) \to \mathbf{D}^+ (\mathcal{A})$.

Question. When are these functors full and/or faithful?

My impression is that the correct definition of bounded derived category is the one denoted by $\mathbf{D}^\mathrm{b} (\mathcal{A})$ above, but it did not occur to me until just now that this might be different from $\mathbf{Q}^{-1} \mathbf{Ch}^\mathrm{b} (\mathcal{A})$. I imagine that if $\mathcal{A}$ has enough projective objects then $\mathbf{Q}^{-1} \mathbf{Ch}^- (\mathcal{A}) \to \mathbf{D}^- (\mathcal{A})$ is an equivalence of categories, and dually if $\mathcal{A}$ has enough injective objects then $\mathbf{Q}^{-1} \mathbf{Ch}^+ (\mathcal{A}) \to \mathbf{D}^+ (\mathcal{A})$ is an equivalence of categories – it should be one of those standard arguments about reducing zigzags using resolutions – but I have not checked, and even if this is correct, it does not answer the question about $\mathbf{D}^\mathrm{b} (\mathcal{A})$. It also leaves open the possibility that there is some impoverished $\mathcal{A}$ where these categories are genuinely different.

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They are always fully faithful, and this doesn't require enough projectives or injectives.

A map in $\mathbf{D}(\mathcal{A})$ from $X$ to $Y$ is represented by a diagram $X\stackrel{s}{\leftarrow}Z\stackrel{f}{\to}Y$ where $s$ is a quasi-isomorphism (call this map $fs^{-1}$), and if $t:Z'\to Z$ is a quasi-isomorphism then $fs^{-1}=(ft)(st)^{-1}$.

But if $X$ (and hence $Z$) has cohomology bounded above, then there is always a quasi-isomorphism $Z'\to Z$ from a bounded above complex: for sufficiently large $n$, let $Z'$ be the "good" truncation $$\tau_{\leq n}Z:=\dots\to Z^{n-2}\to Z^{n-1}\to\ker(d^n)\to0\to0\to\dots$$ of $Z$, and let $t$ be the inclusion map.

So both ways of defining $\mathbf{D}^-(\mathcal{A})$ are "the same".

A dual argument works for $\mathbf{D}^+(\mathcal{A})$.

For $\mathbf{D}^b(\mathcal{A})$ use the first argument followed by the second.

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  • $\begingroup$ Ah yes, it's coming back to me now! Prima facie this argument only shows the functor is full, but this argument plus the calculus of fractions shows faithfulness as well. Very good. $\endgroup$ – Zhen Lin Jul 30 '20 at 4:30

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