# Closing a subcategory under finite colimits by transfinite induction

Let $\mathcal{C}$ be a locally small category with all finite colimits, and let $\mathcal{A}$ be a small full subcategory. I wish to prove the following:

Proposition. There exists a full subcategory $\mathcal{B}$ of $\mathcal{C}$ satisfying these conditions:

• $\mathcal{A} \subseteq \mathcal{B} \subseteq \mathcal{C}$

• Finite colimits exist in $\mathcal{B}$ and are the same as in $\mathcal{C}$.

• Every object of $\mathcal{B}$ is a finite colimit of objects in $\mathcal{A}$.

There is an obvious transfinite process that one can apply here. First, set $\mathcal{B}_0 = \mathcal{A}$. For each ordinal $\alpha$, let $\mathcal{B}_{\alpha + 1}$ be the full subcategory of $\mathcal{C}$ obtained by adding to $\mathcal{B}_\alpha$ any missing finite coproducts or coequalisers. For each limit ordinal $\lambda$, let $\mathcal{B}_\lambda = \bigcup_{\alpha < \lambda} \mathcal{B}_\alpha$.

Question. Why does this transfinite process converge? $\DeclareMathOperator{\ob}{ob}\DeclareMathOperator{\mor}{mor}$

It is easy to obtain an upper bound on the size of $\ob \mathcal{B}_\alpha$: by its very construction, $$\left| \ob \mathcal{B}_{\alpha+1} \right| \le \sum_{n < \aleph_0} \left| \ob \mathcal{B}_\alpha \right|^n + \left| \mor \mathcal{B}_\alpha \right|^2$$ and if we assume $\mathcal{B}_\alpha$ is non-empty, the inequality $\left| \ob \mathcal{B}_\alpha \right| \le \left| \mor \mathcal{B}_\alpha \right|$ then implies $$\left| \ob \mathcal{B}_{\alpha+1} \right| \le \sup \left\lbrace \left| \mor \mathcal{B}_\alpha \right|^n : n < \aleph_0 \right\rbrace \le \left| \mor \mathcal{B}_\alpha \right|^{\aleph_0}$$ which, while elegant, is not too useful since we can't bound $\left| \mor \mathcal{B}_{\alpha+1} \right|$ in terms of $\left| \ob \mathcal{B}_{\alpha+1} \right|$. It is conceivable that $\mathcal{C}$ and $\mathcal{A}$ are chosen in such a way that at each step $\alpha$, an object is added which vastly increases the size of $\mor \mathcal{B}_{\alpha+1}$.

Nonetheless, I'm fairly confident the proposition is true. Adámek and Rosický even say that "it is clear" [LPAC, p. 17] (albeit with extra hypotheses on $\mathcal{A}$ and $\mathcal{C}$) and offer no proof.

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Don't we have $\mathcal B_{\omega + 1} = \mathcal B_\omega$? For if $A_1, \ldots, A_k$ is a finite collection of objects of $\mathcal B_\omega$, then they are in some $\mathcal B_n$, so their coproduct is an object of $\mathcal B_{n+1}$, analogously for the coequalizers ... – martini Aug 23 '12 at 17:29
Ah, that works. And I guess it generalises if I replace "finite" with any other regular cardinal... – Zhen Lin Aug 24 '12 at 1:37
Yes, it should ... – martini Aug 24 '12 at 10:43
While I'm sure it's not the case here, I recall talking to my supervisor before he went to a large cardinals & homotopy theory conference. I was surprised to hear that very large cardinals such as Vopenka cardinals are used in homotopy theory. He then explained that in order to show that such and such categories are closed under such and such you need a transfinite induction, and to guarantee it works you apparently have to use some very large cardinals. – Asaf Karagila Jan 30 '13 at 1:22

[W]e have $\mathcal B_{\omega+1} = \mathcal B_\omega$. For if $A_1, \ldots, A_k$ is a finite collection of objects of $\mathcal B_\omega$, then they are in some $\mathcal B_n$, so their coproduct is an object of $\mathcal B_{n+1}$, analogously for the coequalizers.