On why categories with coseparating sets and intersections have initial objects [Proof Explanation]

Question

I'm trying to understand the proof of the following result from Category Theory in Context, p. 148:

Lemma 4.6.11. Suppose $C$ is locally small, complete, has a small coseparating set $\Phi$, and has the property that every collection of subobjects has an intersection. Then $C$ has an initial object.

The main idea is to take the product $p = \prod_\Phi k$ of the objects in the coseparating set, and then take the intersection $i$ of all subobjects of $p$, which we ought to see is initial. Shortly after the author introduces a canonical map

$$ c \rightarrowtail \prod_{k \in \Phi}k^{C(c,k)} \tag{1} $$

which is claimed to be a mono since $\Phi$ is a coseparating set, and a map

$$ \prod_{k \in \Phi}k \longrightarrow \prod_{k \in \Phi}k^{C(c,k)} \tag{2} $$

defined as 'the product over $k \in \Phi$ of the maps $\Delta: k → k^{C(c,k)}$ defined to be the identity on each component of the power $k^{C(c,k)}$'. Finally, the pullback of these maps $p_c$ is claimed to be a subobject of $p = \prod_\Phi k$, hence inducing a map $i \to p_c$ and composing with the pullback map to $c$, we get an arrow $i \to c$. Finally, the author claims that

'There cannot be more than one arrow from $i$ to $c$ because the equalizer of two distinct such would define a smaller subobject of $p$, contradicting minimality of the intersection $i$. Thus, we conclude that $i$ is an initial object.'

I have several questions:

• how is exactly the map $(2)$ constructed? I have not quite understood what taking the product of $\Delta$ maps means. Surely by the universal property of the product it suffices to give arrows $\prod_k \Phi \to k'$ for each $f \in C(c,k')$ and $k' \in \Phi$.

• why is the pullback cone leg $p_c \to p$ a monomorphism? Is this a specific result or am I missing a general property of pullbacks?

• and finally, I have not quite understood the quoted paragraph. I could take it at face value, but I would really appreciate any comments that can enlighten what is really going on in the background, since it feels like this result has a lot of information compressed in few sentences.

Thanks in advance!

Answer 1

First, just to be clear¹, $i$ and a few other objects/arrows are being used in an overloaded manner in this text. $i$ is a subobject of $\prod_{k\in\Phi}k$ which means its an equivalence class of monomorphisms into $\prod_{k\in\Phi}k$. In a common use of overloading, we let a representative of the equivalence class stand for the whole equivalence class, i.e. we treat $i$ as a specific monomorphism. $i$ is also being treated as an object, namely the domain of this representative monomorphism.

Given arrows $f:A\to B$ and $g:C \to D$ we have the arrow $f\times g:A\times C\to B\times D$. $(2)$ is just the $\Phi$-ary version of this for the maps $\Delta_k:k\to k^{\mathcal C(c,k)}$.

The pullback of a monomorphism is a monomorphism. This is a general result. That is, in the pullback diagram $$\require{AMScd}\begin{CD}A\times_{f,g}B@>p>>A\\@VqVV@VVfV\\B@>>g>C\end{CD}$$ if $f$ is a monomorphism, then so is $q$. (And, of course, symmetrically for $g$ and $p$.) You should prove this.

$i$ is defined as the intersection of the (potentially proper class of) all subobjects of $\prod_{k\in\Phi}k$. The subobjects form a preordered set², and intersection is just the meet (product) in that preordered set. So, like any product, that means there's a projection from $i$ to any of its components which includes all subobjects of $\prod_{k\in\Phi}k$. An equalizer of two arrows $f,g:i\to c$ is an arrow into $i$, and, like any equalizer, is a monomorphism, call it $e$. By composition with $i$, this gives us a (representative of a) subobject of $\prod_{k\in\Phi}k$. Since $i\circ e$ is a subobject that factors through $i$, then $i\circ e\lesssim i$ as subobjects, but, by definition of $i$, $i\lesssim i\circ e$, so $i$ and $i\circ e$ are equivalent as subobjects. This implies that there is some isomorphism, $r$, such that $i\circ e \circ r=i$, but since $i$ is a monomorphism, this means $e\circ r = id$. Finally, we have $f\circ e = g\circ e\iff f\circ e\circ r = g\circ e\circ r\iff f = g$.

¹ I don't think you're confused about this, but it is definitely a potentially confusing aspect of this text.

² Well, preordered class. The difference doesn't really matter here.