nLab internally projective object

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Idea

An object of a category (usually a topos or pretopos) is internally projective if it satisfies the internalization of the condition on a projective object.

Definition

An object EE of a topos 𝒯\mathcal{T} is called internally projective if the internal hom/exponential object functor

(βˆ’) E:𝒯→𝒯 (-)^E \colon \mathcal{T} \to \mathcal{T}

preserves epimorphisms.

Equivalent characterizations

Some of these facts, and more, can be found stated in Exercises 15 and 16 in Chapter IV of Sheaves in Geometry and Logic.

Theorem

EE is internally projective if and only if the dependent product functor (the right adjoint to pullback)

Ξ  E:𝒯/E→𝒯 \Pi_E \colon \mathcal{T}/E \to \mathcal{T}

preserves epimorphisms.

Proof

The functor (βˆ’) E(-)^E is the composite Ξ  E∘E *\Pi_E \circ E^*, and pullback E *:𝒯→𝒯/EE^*\colon \mathcal{T} \to \mathcal{T}/E preserves epis, so if Ξ  E\Pi_E preserves epis then so does (βˆ’) E(-)^E.

Conversely, for any f:Aβ†’Bf:A\to B in 𝒯/E\mathcal{T}/E, we have a pair of pullback squares:

Ξ  EA β†’ A E ↓ ↓ Ξ  EB β†’ B E ↓ ↓ 1 β†’id E E E\array{ \Pi_E A & \to & A^E \\ \downarrow & & \downarrow\\ \Pi_E B & \to & B^E \\ \downarrow & & \downarrow\\ 1 & \xrightarrow{id_E} & E^E }

The lower square and outer rectangle are easily seen to be pullbacks, hence so is the upper square. Since epimorphisms are closed under pullback, the other implication follows.

Lemma

If EE is internally projective in 𝒯\mathcal{T}, then I *EI^* E is internally projective in 𝒯/I\mathcal{T}/I.

Proof

This proof was given by Alex Simpson here. Let q:Bβ†’Aq:B\to A be an epimorphism from v:Bβ†’Iv:B\to I to u:Aβ†’Iu:A\to I. Then the exponential u I *Eu^{I^*E} in 𝒯/I\mathcal{T}/I is the equalizer of

I×A E→pr 2A E→uI E I\times A^E \xrightarrow{pr_2} A^E \xrightarrow{u} I^E

and

I×A E→pr 1I→constI E I\times A^E \xrightarrow{pr_1} I \xrightarrow{const} I^E

equipped with the obvious projection to II. Now we have a pullback square

v I *E β†’ IΓ—B E ↓ ↓ u I *E β†’ IΓ—A E\array{ v^{I^*E} & \to & I\times B^E \\ \downarrow & & \downarrow \\ u^{I^*E} & \to & I\times A^E }

in which the right-hand map is epi since EE is internally projective; hence so is the left-hand map.

Theorem

EE is internally projective if and only if the statement β€œEE is projective” is true in the stack semantics of 𝒯\mathcal{T}.

Proof

By definition, truth of β€œEE is projective” in the stack semantics means that for any II and any epimorphism Aβ†’IΓ—EA\to I\times E, there exists an epimorphism p:Jβ†’Ip:J\to I and a section of (pΓ—id) *Aβ†’JΓ—E(p\times id)^*A \to J\times E. (We have used the characterization that an object XX is projective just when every epimorphism with codomain XX is split.)

If EE is internally projective, then given an epimorphism Aβ†’IΓ—EA\to I\times E as above, let J=Ξ  pr 1(A)J = \Pi_{pr_1}(A), where pr 1:IΓ—Eβ†’Ipr_1:I\times E \to I is the projection. Since IΓ—EI\times E is internally projective in 𝒯/I\mathcal{T}/I by Lemma , p:Jβ†’Ip:J\to I is an epimorphism. And (pΓ—id) *Aβ†’JΓ—E(p\times id)^*A \to J\times E is split by the counit of the adjunction (pr 1) *⊣Π pr 1(pr_1)^* \dashv \Pi_{pr_1}.

Conversely, suppose the above condition holds, and let e:Bβ†’Ae:B\to A be an epimorphism in 𝒯/E\mathcal{T}/E. Let I=Ξ  E(A)I = \Pi_E(A), and let CC be the pullback of ee along the counit Ξ  E(A)Γ—Eβ†’A\Pi_E(A) \times E\to A. Then Ξ  E(B)\Pi_E(B) is equivalently Ξ  pr 1(C)\Pi_{pr_1}(C), where pr 1:IΓ—Eβ†’Ipr_1:I \times E \to I is the projection.

Moreover, C→I×EC \to I\times E is epi, so by assumption there is an epi p:J→Ip:J\to I such that (p×id) *C→J×E(p\times id)^*C \to J\times E is split. Since pullback along epis reflects epis, it suffices to show that p *Π pr 1(C)p^* \Pi_{pr_1}(C) is split. However, we have a pullback square

JΓ—E β†’ IΓ—E pr 1↓ ↓ pr 1 J β†’ I \array{ J\times E & \to & I\times E \\ ^{pr_1}\downarrow & & \downarrow^{pr_1} \\ J & \to & I }

so by the Beck-Chevalley condition, p *Ξ  pr 1(C)p^* \Pi_{pr_1}(C) is equivalently Ξ  pr 1(pΓ—id) *C\Pi_{pr_1}(p\times id)^* C. But (pΓ—id) *C(p\times id)^*C is split, and all functors preserve split epis.

Note that Lemma D4.5.3(iii) of Sketches of an Elephant is the special case of the above stack-semantics version of internal projectivity when I=1I=1. This is insufficient for the implication (iii)β‡’\Rightarrow(ii) of that lemma to hold, since if so, then every projective object would be internally projective, which as we show below is not the case.

Projectivity versus internal projectivity

If the terminal object in 𝒯\mathcal{T} is projective, then every internally projective object is projective. In the converse direction,

Proposition

If 𝒯\mathcal{T} has enough projectives and projectives are closed under binary products, then every projective object is internally projective. (In particular, if all objects of 𝒯\mathcal{T} are projective then all objects are internally projective.)

Proof

Let PP be a projective object. To show that e P:E P→B Pe^P \colon E^P \to B^P is epic whenever e:E→Be \colon E \to B is epic, choose an epi ϕ:P′→B P\phi \colon P' \to B^P where P′P' is projective (using the assumption of enough projectives). Since P×P′P \times P' is projective, there exists a lift through ee of the horizontal composite as shown:

E ↓ e Pβ€²Γ—P →ϕ×1 B PΓ—P β†’eval B;\array{ & & & & E \\ & & & & \downarrow ^\mathrlap{e} \\ P' \times P & \underset{\phi \times 1}{\to} & B^P \times P & \underset{eval}{\to} & B; }

this, by currying, provides a lift of Ο•:Pβ€²β†’B P\phi \colon P' \to B^P through e Pe^P. Since Ο•\phi is epic, this immediately implies e Pe^P is epic, as desired.

Remark

Proposition may fail without the assumption that projective objects are closed under binary products. An example is given here.

Remark

The internal axiom of choice (that is, the axiom of choice interpreted in the internal logic of the topos) is equivalent to the statement that every object is internally projective. This is strictly weaker than the β€œexternal” axiom of choice that every epimorphism in the topos is split.

Corollary

In a presheaf topos Set C opSet^{C^{op}}, if CC has binary products, then every projective presheaf is internally projective.

Proof

Representables, and arbitrary coproducts of representables, are projective, and every presheaf is covered by some coproduct of representables. This implies that projective presheaves are precisely retracts of coproducts of representables. Under the assumption that CC has binary products, coproducts of representables, and also their retracts, are also closed under binary products. Thus projective presheaves are closed under binary products. Now apply Proposition .

Discussion

Last revised on March 30, 2023 at 14:13:40. See the history of this page for a list of all contributions to it.