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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.

The answer to the question B is just a formal consequence(though, I don't think that this is indeed internal Hom for arbitrary category $C$, for category of modules over a ring it is): forgetful functor from abelian groups to sets is fully faithful and commutes with filtered colimits, so arrows in the last diagram are isomorphisms already on the level of complexes.

Edit: Let me add some details on the last argument. I will prove that these are actually isomorphisms of complexes. First, let's prove it for $Hom^0$. Applying forgetful functor $AB\to Sets$ we obtain the diagram from the part A because this forgetful functor commutes with filtered colimits. As it is faithful, it follows that arrows in the part B are indeed isomorphisms of abelian groups. Next, we get the same for whole complexes repeating this argument for $X,Y[i]$.

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.

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 clases of left fractions. Now, Ore conditions guarantee that these two sets of equivalence classes coincide, which is precisely the isomorphism you are looking for.

The answer to the question B is just a formal consequence(though, I don't think that this is indeed internal Hom for arbitrary category $C$, for category of modules over a ring it is): forgetful functor from abelian groups to sets is fully faithful and commutes with filtered colimits, so arrows in the last diagram are isomoprhisms already on the level of complexes.

Edit: Let me add some deatils on the last argument. I will prove that these are actually isomorphisms of complexes. First, let's prove it for $Hom^0$. Applying forgetful functor $AB\to Sets$ we obtain the diagram from the part A because this forgetful functor commutes with filtered colimits. As it is faithul, it follows that arrows in the part B are indeed isomorphisms of abelian groups. Next, we get the same for whole complexes repeating this argument for $X,Y[i]$.

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.

The answer to the question B is just a formal consequence(though, I don't think that this is indeed internal Hom for arbitrary category $C$, for category of modules over a ring it is): forgetful functor from abelian groups to sets is fully faithful and commutes with filtered colimits, so arrows in the last diagram are isomorphisms already on the level of complexes.

Edit: Let me add some details on the last argument. I will prove that these are actually isomorphisms of complexes. First, let's prove it for $Hom^0$. Applying forgetful functor $AB\to Sets$ we obtain the diagram from the part A because this forgetful functor commutes with filtered colimits. As it is faithful, it follows that arrows in the part B are indeed isomorphisms of abelian groups. Next, we get the same for whole complexes repeating this argument for $X,Y[i]$.

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 clases of left fractions. Now, Ore conditions guarantee that these two sets of equivalence classes coincide, which is precisely the isomorphism you are looking for.

The answer to the question B is just a formal consequence(though, I don't think that this is indeed internal Hom for arbitrary category $C$, for category of modules over a ring it is): forgetful functor from abelian groups to sets is fully faithful and commutes with filtered colimits, so arrows in the last diagram are isomoprhisms already on the level of complexes.

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 clases of left fractions. Now, Ore conditions guarantee that these two sets of equivalence classes coincide, which is precisely the isomorphism you are looking for.

The answer to the question B is just a formal consequence(though, I don't think that this is indeed internal Hom for arbitrary category $C$, for category of modules over a ring it is): forgetful functor from abelian groups to sets is fully faithful and commutes with filtered colimits, so arrows in the last diagram are isomoprhisms already on the level of complexes.

Edit: Let me add some deatils on the last argument. I will prove that these are actually isomorphisms of complexes. First, let's prove it for $Hom^0$. Applying forgetful functor $AB\to Sets$ we obtain the diagram from the part A because this forgetful functor commutes with filtered colimits. As it is faithul, it follows that arrows in the part B are indeed isomorphisms of abelian groups. Next, we get the same for whole complexes repeating this argument for $X,Y[i]$.