# Bounded chain complexes and the bounded derived category

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.

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.

• 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. – Zhen Lin Jul 30 '20 at 4:30