# Regular Projective Objects in the Exact Completion of a Finitely Complete Category

Everything I'm talking about is contained in the following:

• Carboni & Celia Magno, The free exact category on a left exact one;

• Carboni, Some free constructions in realizability and proof theory.

If $$C$$ is a finitely complete category, it can be embedded into an exact category $$Ex(C)$$, in such a way that this embedding $$C\hookrightarrow Ex(C)$$ preserves finte limits and is initial among all the possible finite limit preserving functors $$C \to D$$ where $$D$$ is an exact category.

The category $$Ex(C)$$ is defined as follows [$$(1)-(2)-(2')-(3)-(4)$$].

$$(1)$$ An object is a pseudo-equivalence relation $$R \to X\times X$$ in $$C$$. A pseudo equivalence relation is almost an equivalence relation, with the only difference that it is not required to be a monomorphism (in particular, if a pseudo-equivalence relation has a regular epi-mono factorization then its image is an equivalence relation).

$$(2)$$ An arrow between two pseudo-equivalence relations is an arrow of $$C$$ between their supports that agrees with the pseudo-equivalence relations. For instance, if $$R \to X\times X$$ and $$S \to Y\times Y$$ are pseudo-equivalence relations and $$f$$ is an arrow $$X\to Y$$, then $$f$$ is an arrow of $$Ex(C)$$ iff:

for every arrow $$x=\langle x_1,x_2\rangle \colon I \to X\times X$$, if $$x$$ factors through $$R$$ then the arrow $$(f\times f)x=\langle fx_1,fx_2\rangle \colon I \to Y\times Y$$ factors through $$S$$ (if we were talking about sets, this would mean that, whenever $$(x_1,x_2)\in R$$ then $$(fx_1,fx_2)\in S$$),

or, equivalently, iff:

there is an arrow $$R \xrightarrow{f'} S$$ (not necessarily unique) making the obvious square commute.

$$(2')$$ Actually, we consider equivalent two such arrow $$f,g \colon X \to Y$$ such that for every $$x \colon I \to X$$, the arrow $$\langle fx,gx\rangle \colon I \to Y \times Y$$ factors through $$S$$ (again, set-theoretically this means that $$(fx,gx) \in S$$ for every $$x\in X$$). This happens precisely when $$\langle f,g\rangle$$ factors through $$S$$. Hence an arrow of $$Ex(C)$$ is an equivalence class, modulo this relation, of equivalence relation preserving parallel arrows.

$$(3)$$ One can verify that this equivalence relation between arrows agrees with the composition of $$C$$, hence we can define the composition in $$Ex(C)$$ of two classes as the class of the composition in $$C$$ of their representatives.

$$(4)$$ The embedding $$C \hookrightarrow Ex(C)$$ sends every object $$X$$ of $$C$$ to the trivial (pseudo-)equivalence relation $$\langle 1_X,1_X\rangle \colon X \to X\times X$$ and every arrow $$X \to Y$$ to itself. Obviously this defines an embedding and one can verify that it preserves finite limits.

One can prove that $$Ex(C)$$ is exact. In particular, the regular-epi mono factorization is obtained as follows: if $$[f]$$ is an arrow $$(X,R) \to (Y,S)$$, then its image is the arrow $$(X,(f\times f)^*S)\to(Y,S)$$ again represented by $$f$$, and the regular epimorphism $$(X,R) \to (X,(f\times f)^*S)$$ is the arrow represented by the identity $$1_X$$ (indeed, by the universal property of $$(f\times f)^*S$$, there is an arrow $$R \to (f\times f)^*S$$ making the obvious square commute - just consider the arrows $$R \to X\times X$$ and $$f'$$ in order to get it). Hence we know how the regular epimorphisms look like.

One can characterize the exact categories obtained through this procedure. Indeed if $$D$$ is an exact category with enough projective (means regular projective) objects and its full subcategory $$P$$ of its projective objects is closed under finite limits (that is, finitely complete) then $$D \simeq Ex(P)$$. Viceversa, if $$C$$ is a finitely complete category, then the inclusion of $$C$$ into $$Ex(C)$$ through the embedding $$C \hookrightarrow Ex(C)$$ verifies the following properties:

$$(a)$$ it is finitely complete (this is trivial because of $$(4)$$);

$$(b)$$ its objects are projective in $$Ex(C)$$ and the projective objects are enough;

$$(c)$$ its objects, up to isomorphism of $$Ex(C)$$, are precisely the projective objects of $$Ex(C)$$, that is, every projective object of $$Ex(C)$$ is in the essential image of $$C \hookrightarrow Ex(C)$$.

The only thing I can't prove is $$\boldsymbol{(c)}$$. Can someone help me working it out?

Maybe the proof that $$Ex(C)$$ has enough projectives can help: let $$(X,R)$$ be a pseudo equivalence relation over the object $$X$$ of $$C$$. After having proved that $$(X,\langle 1_X,1_X\rangle)$$ is projective, observe that the identity $$1_X$$ represents a regular epimorphism $$(X,\langle 1_X,1_X\rangle) \to (X,R)$$. I was trying to see if this is actually an isomorphism (i.e. a monomorphism - it's enough in such a category) assuming that $$(X,R)$$ is projective, but this doesn't seem to work.

Solution by Fabio P.

Let $$(X,R)$$ be a projective object of $$Ex(\mathcal{C})$$. As we know, the identity $$1_X$$ represents a regular epimorphism $$(X,\langle 1_X,1_X\rangle)\xrightarrow{c}(X,R)$$. It is a wrong idea to try to prove that $$c$$ is an isomorphism. Indeed, $$(X,R)$$ turns out to be isomorphic to an object in the image of $$\mathcal{C}\hookrightarrow Ex(\mathcal{C})$$ which is not necessarily $$(X,\langle 1_X,1_X\rangle)$$.

As $$(X,R)$$ is projective, there is an arrow $$X \xrightarrow{s}X$$ of $$\mathcal{C}$$, representing a section $$(X,R)\to (X,\langle 1_X,1_X\rangle)$$ of $$c$$.

Let $$(E,S)\xrightarrow{[e]}(X,\langle 1_X,1_X\rangle)$$ be the equaliser of the couple $$([s]c,[1_X])$$ of arrows $$(X,\langle 1_X,1_X\rangle)\to(X,\langle 1_X,1_X\rangle)$$. Then $$[e]$$ is a monomorphism of $$Ex(\mathcal{C})$$, hence, up to precomposing by an isomorphism, we can assume (look at the exhibition we gave of the regular epi-mono factorisation in $$Ex(\mathcal{C})$$) that: $$S=(e\times e)^*\langle 1_X,1_X\rangle=\langle e^*1_X,e^*1_X\rangle=\langle 1_E,1_E\rangle.$$ Observe that the arrow $$[s]$$ equalises the pair $$([s]c,[1_X])$$. Hence, by the universal property of the arrow $$[e]$$, there is unique an arrow $$(X,R)\xrightarrow{r}(E,\langle 1_E,1_E\rangle)$$ such that $$[e]r=[s]$$.

We are done if $$r$$ is an isomorphism. This is indeed true, since:

• it is the case that $$(c[e])r=c[s]=[1_X]=1_{(X,R)}$$;

• it is the case that $$[e]r(c[e])=[s]c[e]=[e]$$, hence $$r(c[e])=1_{(E,\langle 1_E,1_E\rangle)};$$

that is, $$c[e]$$ is the inverse of $$r$$.