# Defining hom spaces in the derived category as limits of hom spaces in the homotopy category

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: $$\lim_{X' \underset{qis}\to X} Hom^{\bullet}(X',Y) \widetilde\to \lim_{X' \underset{qis}\to X,Y \underset{qis}\to Y'} Hom^{\bullet}(X',Y') \widetilde\leftarrow \lim_{Y \underset{qis}\to Y'}Hom^{\bullet}(X,Y')$$

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).
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, $$X=A, Y=A\xrightarrow{1} A$$ 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.
• @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. – Denis-Charles Cisinski Jan 31 '19 at 15:48