Questions: Derived Categories and Derived Equivalences

Question 1

The homotopy category K(𝒜) identifies chain maps up to homotopy equivalence. The derived category D(𝒜) makes an additional identification. What is it?

• A. D(𝒜) further identifies all complexes with their total cohomology, collapsing each complex to a graded group
• B. D(𝒜) formally inverts quasi-isomorphisms, so complexes with the same homology groups become isomorphic even if they are not homotopy equivalent
• C. D(𝒜) identifies complexes up to exact functor equivalence, splitting all short exact sequences
• D. D(𝒜) collapses all bounded complexes to their degree-0 cohomology, making it equivalent to 𝒜 itself

Answer: (B)

K(𝒜) only identifies chain maps modulo homotopy; two complexes can be quasi-isomorphic in K(𝒜) yet remain non-isomorphic objects. D(𝒜) is obtained by localizing K(𝒜) at the class of quasi-isomorphisms — formally inverting them — so any quasi-isomorphism becomes an isomorphism in D(𝒜). This means complexes that encode the same homological data (same homology groups) are genuinely isomorphic objects in D(𝒜). This is strictly finer than homotopy equivalence: quasi-isomorphic complexes need not be homotopy equivalent.

Question 2

In the derived category D(𝒜), how does the classical Ext^n(A, B) appear?

• A. As the n-th cohomology of the internal Hom complex Hom(A, B)
• B. As the set of homotopy classes of chain maps from A to B of degree n in K(𝒜)
• C. As Hom_{D(𝒜)}(A, B[n]), the set of morphisms in D(𝒜) from A to the n-fold shift of B
• D. As a derived functor that must still be computed externally via a projective resolution of A

Answer: (C)

One of the central payoffs of the derived category is the isomorphism Ext^n(A, B) ≅ Hom_{D(𝒜)}(A, B[n]), where B[n] is the complex B shifted n degrees. Derived functors, which classically required external resolution computations, become representable as morphisms within D(𝒜) itself. This is why derived categories are described as the 'natural home' of homological algebra: Ext is no longer a separate gadget but a Hom set in the correct category. Option D describes the classical construction that the derived category supersedes.

Question 3

The derived category D(𝒜) equals the homotopy category K(𝒜) whenever the abelian category 𝒜 has enough injectives, because in that case most quasi-isomorphism is also a homotopy equivalence.

• T. True
• F. False

Answer: False

Having enough injectives does not make quasi-isomorphisms into homotopy equivalences. Even in categories with enough injectives (such as modules over a ring), there exist quasi-isomorphic complexes that are not homotopy equivalent. The derived category adds new isomorphisms beyond those in K(𝒜) regardless of whether injectives exist. What enough injectives enables is the use of injective resolutions to compute derived functors — but this is about computational technique, not about collapsing D(𝒜) into K(𝒜).

Question 4

In D(𝒜), an object A (viewed as a complex concentrated in degree 0) is isomorphic to any of its injective resolutions.

• T. True
• F. False

Answer: True

An injective resolution of A is a complex I• that is quasi-isomorphic to A (the augmentation map A → I• induces isomorphisms on all homology groups). Since D(𝒜) is obtained by inverting quasi-isomorphisms, A and I• become isomorphic in D(𝒜). This is the fundamental reason derived functors are natural in the derived category: RF(A) = F(I•) is not an external gadget but the image of A's isomorphic copy I• under F, making resolutions internal to the categorical structure.

Question 5

Explain the difference between a quasi-isomorphism and a homotopy equivalence of chain complexes, and why D(𝒜) inverts quasi-isomorphisms rather than just homotopy equivalences.

Model answer: A homotopy equivalence is a chain map f: A→B with a chain homotopy inverse g: B→A such that gf and fg are chain homotopic to identity. A quasi-isomorphism is any chain map inducing isomorphisms on all homology groups, but it may not have any chain map as an inverse — it is a strictly weaker notion. Every homotopy equivalence is a quasi-isomorphism, but not conversely. D(𝒜) inverts quasi-isomorphisms because the correct notion of 'same object' in homological algebra is 'same homology,' not 'homotopy equivalent complex.' By inverting quasi-isomorphisms, D(𝒜) identifies A with all its resolutions, making derived functors representable as morphisms. Stopping at homotopy equivalences would leave distinct the objects and their resolutions — preventing Ext from being a Hom set in the category.

A concrete example: the complex 0 → ℤ →×2→ ℤ → 0 is quasi-isomorphic to ℤ/2 concentrated in degree 0 (they have the same cohomology), but they are not homotopy equivalent as chain complexes. In D(Ab), they become isomorphic, which is the correct identification for homological purposes.