Questions: Projective Objects and Projective Covers

Question 1

Let P be a projective R-module, e: C ↠ B a surjective module homomorphism, and f: P → B a module map. What does the projective property guarantee?

• A. There exists a surjection g: C ↠ P such that f ∘ g = e
• B. The module C must also be projective whenever P maps into B via a surjection from C
• C. There exists a homomorphism f̃: P → C such that e ∘ f̃ = f, i.e., the map f lifts through the surjection e
• D. The kernel of e is isomorphic to P as an R-module

Answer: (C)

The defining property of a projective object is the lifting property: given any epimorphism e: C ↠ B and any morphism f: P → B, there exists a morphism f̃: P → C making the triangle commute (e ∘ f̃ = f). The map from P to B can always be 'lifted' to a map from P to C passing through the surjection. This says nothing about C being projective, about the kernel of e, or about maps in the reverse direction — it is purely about the existence of a compatible lift.

Question 2

Over a general commutative ring R (not necessarily a principal ideal domain), which statement about projective and free modules is correct?

• A. Projective and free are synonymous in any module category — 'projective' is just an abstract name for 'free'
• B. Every free module is projective, but projective modules need not be free — they are precisely the direct summands of free modules
• C. Every projective module is free, but not all free modules are projective over non-commutative rings
• D. Projective modules are the quotients of free modules, making them strictly weaker than free modules

Answer: (B)

Every free module is projective (an explicit lift can always be constructed using the basis). But over general rings, projective-but-not-free modules exist: M is projective if and only if there exists N such that M ⊕ N is free — M is a direct summand of a free module, but need not be free itself. Over a PID like ℤ, every projective module is free. The Serre-Swan theorem gives geometric content to the distinction: finitely generated projective modules over C(X) correspond to vector bundles over X, and non-trivial vector bundles are projective-but-not-free.

Question 3

Projective objects and injective objects are categorical duals: the definition of each is obtained from the other by reversing all arrows, replacing surjections with injections, and lifts with extensions.

• T. True
• F. False

Answer: True

An injective object I satisfies: for every monomorphism m: A ↪ B and any map f: A → I, there exists an extension f̃: B → I with f̃ ∘ m = f. Comparing with the projective definition (lifting through epimorphisms) reveals the precise categorical duality — the direction of the diagram's 'given' morphisms reverses, epimorphisms become monomorphisms, and lifting (factoring P → B through C ↠ B) becomes extending (factoring A → I through A ↪ B). Passing to the opposite category converts projective definitions into injective ones and vice versa.

Question 4

Nearly every abelian category has enough projective objects, so projective resolutions and minimal projective covers are universally available tools in homological algebra.

• T. True
• F. False

Answer: False

Projective covers — and even projective objects — do not exist in every abelian category. The category of sheaves on a topological space is a standard example of an abelian category that in general lacks enough projectives (and often lacks projective covers entirely). Projective resolutions are available in module categories over rings (which have enough projectives), but this is a special property, not a universal feature of abelian categories. When projective covers do not exist, one typically works with injective resolutions instead, or accepts non-minimal projective resolutions.

Question 5

Explain why the projective lifting property makes projective modules the natural building blocks for resolutions used in homological algebra.

Model answer: To compute derived functors (like Tor and Ext), you need to replace a module M with a resolution consisting of objects on which functors behave simply — specifically, objects where applying Hom(−, N) or (−) ⊗ N preserves exactness. Projective modules are precisely those on which Hom(P, −) preserves epimorphisms (the defining lifting property), meaning exact sequences remain exact after applying Hom(P, −). By replacing M with a projective resolution P₀ ← P₁ ← P₂ ← ⋯ and applying the functor of interest, you can read off the derived functors from the cohomology of the resulting complex. The lifting property ensures that morphisms between modules can always be compared through projective objects, making projective resolutions the canonical tool for measuring how far a functor deviates from exactness.

This is why projective covers and minimal projective resolutions (where each Pᵢ is as small as possible, with a superfluous kernel) are particularly useful: they give the most economical way to resolve M, producing the sharpest invariants.