category theory

# Contents

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

$(-)^E \colon \mathcal{T} \to \mathcal{T}$

preserves epimorphisms.

## Equivalent characterizations

###### Theorem

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

$\Pi_E \colon \mathcal{T}/E \to \mathcal{T}$

preserves epimorphisms.

###### Proof

The functor $(-)^E$ is the composite $\Pi_E \circ E^*$, and pullback $E^*\colon \mathcal{T} \to \mathcal{T}/E$ preserves epis, so if $\Pi_E$ preserves epis then so does $(-)^E$.

Conversely, for any $f:A\to B$ in $\mathcal{T}/E$, we have a pair of pullback squares:

$\array{ \Pi_E A & \to & A^E \\ \downarrow & & \downarrow\\ \Pi_E B & \to & B^E \\ \downarrow & & \downarrow\\ 1 & \xrightarrow{id_A} & A^A }$

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

###### Proof

This proof was given by Alex Simpson here. Let $q:B\to A$ be an epimorphism from $v:B\to I$ to $u:A\to I$. Then the exponential $u^{I^*E}$ in $\mathcal{T}/I$ is the equalizer of

$I\times A^E \xrightarrow{pr_2} A^E \xrightarrow{u} I^E$

and

$I\times A^E \xrightarrow{pr_1} I \xrightarrow{const} I^E$

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

$\array{ v^{I^*E} & \to & I\times B^E \\ \downarrow & & \downarrow \\ u^{I^*E} & \to & I\times A^E }$

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

###### Theorem

$E$ is internally projective if and only if the statement “$E$ is projective” is true in the stack semantics of $\mathcal{T}$.

###### Proof

By definition, truth of “$E$ is projective” in the stack semantics means that for any $I$ and any epimorphism $A\to I\times E$, there exists an epimorphism $p:J\to I$ and a section of $(p\times id)^*A \to J\times E$. (We have used the characterization that an object $X$ is projective just when every epimorphism with codomain $X$ is split.)

If $E$ is internally projective, then given an epimorphism $A\to I\times E$ as above, let $J = \Pi_{pr_1}(A)$, where $pr_1:I\times E \to I$ is the projection. Since $I\times E$ is internally projective in $\mathcal{T}/I$ by Lemma 1, $p:J\to I$ is an epimorphism. And $(p\times id)^*A \to J\times E$ is split by the counit of the adjunction $(pr_1)^* \dashv \Pi_{pr_1}$.

Conversely, suppose the above condition holds, and let $e:B\to A$ be an epimorphism in $\mathcal{T}/E$. Let $I = \Pi_E(A)$, and let $C$ be the pullback of $e$ along the counit $\Pi_E(A) \times E\to A$. Then $\Pi_E(B)$ is equivalently $\Pi_{pr_1}(C)$, where $pr_1:I \times E \to I$ is the projection.

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

$\array{ J\times E & \to & I\times E \\ ^{pr_1}\downarrow & & \downarrow^{pr_1} \\ J & \to & I }$

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

Note that Lemma 4.5.3(iii) of Sketches of an Elephant is the special case of the above stack-semantics version of internal projectivity when $I=1$. This is insufficient for the implication (iii)$\Rightarrow$(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 $\mathcal{T}$ is projective, then every internally projective object is projective. In the converse direction,

###### Proposition

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

###### Proof

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

$\array{ & & & & E \\ & & & & \downarrow ^\mathrlap{e} \\ P' \times P & \underset{\phi \times 1}{\to} & B^P \times P & \underset{eval}{\to} & B; }$

this, by currying, provides a lift of $\phi \colon P' \to B^P$ through $e^P$. Since $\phi$ is epic, this immediately implies $e^P$ is epic, as desired.

###### Remark

Proposition 1 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 “external” axiom of choice that every epimorphism in the topos is split.

###### Corollary

In a presheaf topos $Set^{C^{op}}$, if $C$ 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 $C$ 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 1.

## Discussion

Revised on August 24, 2013 08:52:09 by Mike Shulman (107.194.22.192)