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My questions are triggered by Borceux, Vol 1, Proposition 4.5.6. The relevant part of the book is browsable on Google Books, but i'll go ahead and reproduce at least the statement here anyway:

Let $\mathfrak{C}$ be a category with pullbacks and universal coproducts (see section 2.14). Given a family $(G_i)_{i \in I}$ of objects of $\mathfrak{C}$. the following conditions are equivalent:

(1) $(G_i)_{i \in I}$ is a regular family of generators

(2) $(G_i)_{i \in I}$ is a dense family of generators

Since the definition of "universal coproducts" is not browsable in google books, i'll attempt to paraphrase Borceux's definition: For a category to have universal coproducts, whenever $\eta: E \rightarrow \Delta Y$ are the coprojections of a coproduct, and $h: X \rightarrow Y$ any arrow, then the pullback of $\eta$ along $\Delta h$ is another set of coprojections for another coproduct. Here i use $\Delta$ to denote the usual diagonal functor.

So my questions are as follows:

  1. I do not think that the proof printed in Borceux is correct. Does anyone disagree with my criticisms, reproduced below?

  2. I have produced what i think is a valid proof, but it is somewhat long winded (reproduced below). Does anyone have anything better?

  3. Does anyone have any other references in the literature where this result (or a very similar one) is also stated and proved, for comparison's sake?

  4. Is there an alternative, less overloaded, term for the concept of "universal coproducts" as defined in Borceux 2.14 and also (in an apparently weaker form) in Jacobs, 1.5.3. I vaguely remember seeing the same concept documented elsewhere under a different name, but i have forgotten where.

Here are my criticisms concerning Borceux's proof:

  1. Minor misprint; I think $g \circ m$ should instead say $g_m$, where it appears.

  2. At a later stage in the proof, Borceaux says that "... $(H_{f \circ u},(w_f)_f)$ and $(H_{f \circ v},(z_f)_f)$ are coproducts ...", but neither of these are of the right shape (i.e. they are not cocones) let alone (coprojections of a) coproduct, so i dispute this step altogether.

  3. Let's take a simple case where $\mathfrak{C}$ is Sets and $\mathcal{G}$ is the family consisting of just one object - a two element set. Let $C$ also be a two element set. Then $\coprod_f G_i$ is an eight element set. By restricting the kernel pair of $\gamma_C$, you may pick a minimal $X$ (with just 6 elements in it) and $u,v : X \rightarrow \coprod_f G_i$ such that $\gamma_C$ is a co-equaliser in such a way that $H_{f \circ u \circ v}$ is always empty. Thus the central result, that $g \circ u \circ x_f \circ z_f \circ l = g \circ v \circ x_f \circ z_f \circ l$ is vacuously true in this case, and goes no way to showing that $g \circ u = g \circ v$.

  4. I also dispute the idea that we can deduce that $g \circ u = g \circ v$ by considering the cone property of $g_f$ in regards of only parallel pairs of arrows in $\mathcal{G}/C$, and the proof looks only at these scenarios. In the example produced in the previous criticism, the only non-trivial parallel pairs of arrows are not enough to fix $g_f$ in this way.

Here is my attempt at a correct proof. It took quite a bit of headscratching to come up with. I eventually learned that you have to use the universal coproducts quite aggresively to keep chopping up the objects involved into manageable chunks until you can use familar diagram chases to tease out a result. Because the proof is a bit more complicated, I'm going to take the liberty of using a different (and hopefully more familiar) set of symbols and notational style from that in Borceux. The artefacts of the proof are represented on the following diagram in $\mathbb{C}$:

$$ \begin{array}{cccccccc} \mathcal{G}(G) & \xrightarrow{g} & \Gamma_r(s) & \xrightarrow{\hat{\rho}_s} & \Gamma_s(s) & \xrightarrow{\hat{\sigma}{s}} & \mathcal{G}(S) \\ & & \downarrow {\scriptstyle \rho_s} & & \downarrow {\scriptstyle \sigma_s} & & \downarrow {\scriptstyle \eta_s} & \searrow {\scriptstyle s} \\ & & \Gamma_s(t) & \xrightarrow{\tau_t} & P & \overset{s'}{\underset{t'}{\rightrightarrows}} & Q & \overset{\hat{\lambda}}{\longrightarrow} & C \\ & & & \underset{=}{\searrow} & \uparrow {\scriptstyle \tau_t} & & \uparrow {\scriptstyle \eta_t} & \nearrow {\scriptstyle t} \\ & & & & \Gamma_t(t) & \xrightarrow{\hat{\tau}_t} & \mathcal{G}(T) \end{array} $$

  1. Let the regular generating family be represented as a full and faithful functor $\mathcal{G} : \mathcal{G} \rightarrow \mathbb{C}$ (we confuse the functor with its domain for economy)

  2. Take arbitrary $C \in \text{Ob}(\mathbb{C})$ and let $\mathcal{G} \downarrow C$ be the usual comma category, with $\Gamma : (\mathcal{G} \downarrow C) \rightarrow \mathbb{C}$ the canoncal projection and $\Gamma_0 : (\mathcal{G}_0 \downarrow C) \rightarrow \mathbb{C}$ be the corresponding restriction to just objects. Objects of $\mathcal{G} \downarrow C$ will be confused with their underlying arrows $f : \mathcal{G}(G) \rightarrow C$ say, without danger of real confusion.

  3. Take $Q$ the coproduct of $\Gamma_0$ and $\eta : \Gamma_0 \rightarrow \Delta Q$ the coprojection. Let $\lambda : \Gamma \rightarrow \Delta C$ be the canoncal cocone with components $\lambda_f = f$ for all objects $f$ of $\mathcal{G} \downarrow C$. Let $\hat{\lambda} : Q \rightarrow C$ be the transpose along $\eta$ of (the restriction of) $\lambda$.

  4. According to the hypothesis concerning $\mathcal{G}$ being a regular generating family, then $\hat{\lambda}$ is the co-equaliser of a pair of arrows $s', t' : P \rightarrow Q$ say.

  5. Our aim is to show that $\lambda$ is the cocone of a colimit. To that end let $\mu : \Gamma \rightarrow \Delta D$ be any other cocone, and let $\hat{\mu} : Q \rightarrow D$ be the transpose along $\eta$. As long as we can show that this means that $\hat{\mu} \cdot s' = \hat{\mu} \cdot t'$, then a fairly simple appeal to the coequaliser and coproduct already mentioned shows that there is a unique $h : C \rightarrow D$ such that $\Delta h \cdot \lambda = \mu$ and then the overall result follows immediately. The remainder of the proof focusses therefore only on showing that $\hat{\mu} \cdot s' = \hat{\mu} \cdot t'$.

  6. Pullback $\eta$ along $\Delta s'$ to obtain $\Gamma_s$ and arrows $\sigma$ and $\hat{\sigma}$ (as seen in the above diagram). Similarly pullback $\eta$ along $\Delta t'$ to obtain $\Gamma_t$ and arrows $\tau$ and $\hat{\tau}$.

  7. Choose any objects $s : \mathcal{G}(S) \rightarrow C$ and $t : \mathcal{G}(T) \rightarrow C$ of $\mathcal{G} \downarrow C$. Then let $\rho$ be the pullback of $\sigma$ along $\Delta \tau_t : \Delta \Gamma_s(t) \rightarrow \Delta P$ to obtain object $\Gamma_r$ and arrows $\rho$ and $\hat{\rho}$.

  8. By the "universal coproduct" hypothesis, $\Gamma_r$, $\Gamma_t$, and $\Gamma_s$ are coproducts with coprojections $\rho$, $\tau$, and $\sigma$ respectfully.

  9. Take suitable components to obtain the diagram above. The entire diagram commutes. Let $f$ be the entire composite from $\mathcal{G}(G)$ on the left to $C$ on the right. This gives three objects $f$, $s$, and $t$ and two arrows $\hat{\sigma}_s \cdot \hat{\rho}_s \cdot g$ and $\hat{\tau}_t \cdot \rho_s \cdot g$ in the category $\mathcal{G} \downarrow C$. Using the fact that $\mu$is a cocone, we thus learn that the above diagram commutes fully with $\mu_s$ in place of $s$, $\mu_t$ in place of $t$ and $\hat{\mu}$ in place of $\hat{\lambda}$.

  10. This in turns tells us that $\hat{\mu} \cdot s' \cdot \tau_t \cdot \rho_s \cdot g = \hat{\mu} \cdot t' \cdot \tau_t \cdot \rho_s \cdot g$. Considering that $g$, $s$, and $t$ were all arbitrary, $\mathcal{G}$ is a generator set, and $\tau$ and $\rho$ are coprojections, we learn than $\hat{\mu} \cdot s' = \hat{\mu} \cdot t'$, and the proof is complete.

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"Universal" is fairly standard and goes back to SGA 4. I suppose you could say "stable under pullback" instead. –  Zhen Lin May 10 '14 at 1:07

1 Answer 1

up vote 1 down vote accepted

For what it's worth a year later, I just made the same complaints as you and corrected the proof in the same way.

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