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Let $C$ be an abelian category and $K(C)$ the homotopy category of complexes in $C$. I've seen the following claimed in several sources (without proof):

A. The following isomorphisms hold: $$\lim_{X' \underset{qis}\to X} Hom_{K(C)}(X',Y) \widetilde\to \lim_{X' \underset{qis}\to X,Y \underset{qis}\to Y'} Hom_{K(C)}(X',Y') \widetilde\leftarrow \lim_{Y \underset{qis}\to Y'}Hom_{K(C)}(X,Y') $$

The limits (actually colimits in this case) are taken over all quasi-isomorphisms.

Why is this true? And how can one prove this?

Can this statement be upgraded to a statement about the internal Hom bifunctor? Something along the lines of:

B. The following quasi-isomorphisms hold: $$\lim_{X' \underset{qis}\to X} Hom^{\bullet}(X',Y) \widetilde\to \lim_{X' \underset{qis}\to X,Y \underset{qis}\to Y'} Hom^{\bullet}(X',Y') \widetilde\leftarrow \lim_{Y \underset{qis}\to Y'}Hom^{\bullet}(X,Y')$$

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This is actually a standard property of the homotopy category of complexes on which construction of the derived category is based, formulated in unusual way.

For a quasi-isomorphism $s:X'\to X$ and $f\in Hom(X',Y)$ denote by $fs^{-1}$ the image of $f$ in the first colimit(in a second we will justify the choice of notation). By description of the filtered colimit, two images $fs^{-1}$ and $gt^{-1}$(for $t:X''\to X,g\in Hom(X'',Y)$) coincide iff there exists $X'''$ and qisms $p:X'''\to X',q:X'''\to X''$ and a map $r\in Hom(X''',Y)$ such that $sp=tq,fp=gq$. This is exactly the condition for fractions $fs^{-1}$ and $gt^{-1}$ to be equivalent in the right localization by qisms(see e.g. Weibel, 10.3).

Analogously, the third colimit can be identified with the set of equivalence classes of left fractions. Now, Ore conditions guarantee that these two sets of equivalence classes coincide, which is precisely the isomorphism you are looking for.

Edit: As Denis-Charles Cisinski explains, the answer to B is negative. Indeed, let's pick a non-zero object $A$ in $C$ and consider, for instance, $X=A[0], Y=A\xrightarrow{1} A$ where $Y$ is a contractible complex concentrated in degrees $-1,0$. The second and third colimits are zero as the category of quasi-isomorphisms $Y\to Y'$ has a final object $Y\to 0$. But the first colimit is not zero: for $X'=X$ there is a non-zero element $Id_A\in Hom(A,A)=Hom^0(X,Y)$ which survives in the colimit because for a quasi-isomorphism $X'\to X$ the map $(X')^0\to A$ has to be non-zero.

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  • $\begingroup$ You are completely right about the "not internal". It's obviously a bifunctor to the category of complexes of k modulez if C is a k-linear category. I dont understand what you said about the first being a formal consequence. Could you elaborate on why that's true? If it is indeed so than we can get the first by nothing just taking zeroth cohomology and using the fact it commutes with filtered colimits. $\endgroup$ – Saal Hardali Sep 7 '16 at 0:13
  • $\begingroup$ @SaalHardali I tried to elaborate, please let me know if something is unclear. $\endgroup$ – SashaP Sep 7 '16 at 14:27
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    $\begingroup$ @SashaP Your argument about part B are not correct. The colimits in A are indexed by maps up to chain-homotopy equivalences. However, in B, we cannot work up to chain homotopy equivalence (because we precisely do not want to take $H^0$ of the Hom's), so that the colimits are indexed by categories which are not filtered and are thus much more complicated to compute. In fact the isomorphisms suggested in B do not hold unless $C$ is trivial. $\endgroup$ – Denis-Charles Cisinski Jan 31 '19 at 15:48
  • $\begingroup$ @Denis-CharlesCisinski Oh, you're completely right, not sure what I had in mind back then. I've edited the answer. Thank you! $\endgroup$ – SashaP Feb 1 '19 at 5:09

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