# Projective object

In the mathematical field of category theory , projective objects are a generalization of the notion of freedom in algebra.

## definition

An object P of a category C is called projective if there is for every epimorphism and every one such that is. This means that the diagram opposite is commutative. So it is projective if and only if the induced mapping for all epimorphisms${\ displaystyle \ alpha \ colon A \ rightarrow B}$${\ displaystyle f \ colon P \ rightarrow B}$${\ displaystyle f ^ {*} \ colon P \ rightarrow A}$${\ displaystyle \ alpha \ circ f ^ {*} = f}$${\ displaystyle P}$${\ displaystyle \ alpha \ colon A \ rightarrow B}$

${\ displaystyle \ operatorname {Mor} _ {\ mathcal {C}} (P, A) \ ni f ^ {*} \ mapsto \ alpha \ circ f ^ {*} \ in \ operatorname {Mor} _ {\ mathcal {C}} (P, B)}$ is surjective.

## Examples

• Every starting object in a category is projective.
• In the category of sets Me , every object is projective. This is a consequence of the axiom of choice .
• The co- product of projective objects is projective.
• Projective groups are precisely the free groups .

## properties

Is in the category of each object the quotient of a projective object, i. H. if there is an epimorphism for every object in which is projective, it is also said that there are enough projective objects . This property plays a role in the context of derived functors . For example, the category of groups has enough projective objects because each group is the quotient of a free group (representation by generators and relations). ${\ displaystyle C}$${\ displaystyle X \ in \ operatorname {Ob} (C)}$${\ displaystyle P \ rightarrow X}$${\ displaystyle P}$${\ displaystyle C}$

## Projective module

In the category of modules one can say more precisely about projective modules.

The following statements are equivalent for a module . ${\ displaystyle P}$

• ${\ displaystyle P}$ is projective.
• For every epimorphism there is such a thing as that . That is, every target epimorphism is a retraction .${\ displaystyle f \ colon M \ rightarrow P}$${\ displaystyle g \ colon P \ rightarrow M}$${\ displaystyle f \ circ g = \ mathbf {1} _ {P}}$${\ displaystyle P}$
• Every epimorphism falls apart. That is, the direct summand is in .${\ displaystyle f \ colon M \ rightarrow P}$${\ displaystyle \ operatorname {core} (f)}$${\ displaystyle M}$
• ${\ displaystyle P}$ is isomorphic to a direct summand of a free module .
• The functor is exact .${\ displaystyle \ operatorname {Hom} (P, -)}$

The direct sum of a family of modules is projective if and only if each is projective. In particular, every direct summand of a projective module is projective. The product of projective modules is generally by no means projective. For example, it is not projective. ${\ displaystyle (P_ {i} | i \ in I)}$${\ displaystyle P_ {i}}$${\ displaystyle \ mathbb {Z} ^ {\ mathbb {N}}}$

### Dual base lemma

A module is created by . The module is projective if and only if there is a family of homomorphisms from the dual space with: ${\ displaystyle P}$${\ displaystyle (y_ {i} | i \ in I)}$${\ displaystyle P}$${\ displaystyle (f_ {i} | i \ in I)}$${\ displaystyle P ^ {*} \ colon = \ operatorname {Hom} (P, R)}$

1. For each is only for a finite number .${\ displaystyle p \ in P}$${\ displaystyle f_ {i} (p) \ neq 0}$${\ displaystyle i \ in I}$
2. For each is .${\ displaystyle p \ in P}$${\ displaystyle \ textstyle p = \ sum _ {i \ in I} y_ {i} f_ {i} (p)}$

### Consequences from the dual basis lemma

• For each right module there is a left module above the ring . This module is called the too dual module. The module is again a legal module. You have natural homomorphism . Is projective, so is injective.${\ displaystyle P}$${\ displaystyle P ^ {*}: = \ operatorname {Hom} (P, R)}$${\ displaystyle R}$${\ displaystyle P}$${\ displaystyle P ^ {**}: = \ operatorname {Hom} (P ^ {*}, R)}${\ displaystyle {\ begin {aligned} \ Phi (P) \ colon P \ ni p & \ mapsto (\ operatorname {Hom} (P, R) \ ni \ alpha \ mapsto \ alpha (p) \ in R) \ in P ^ {**} \ end {aligned}}}${\ displaystyle P}$${\ displaystyle \ Phi (P)}$
• If projective and finitely generated, then there is an isomorphism. It is said to be reflexive.${\ displaystyle P}$${\ displaystyle \ Phi (P)}$${\ displaystyle P}$