# Verifying a Construction Satisfies the $\Omega$-axiom.

It's in the topos $$\mathbf{Bn}(I)$$ of bundles over a set $$I$$. Goldblatt asks the reader to verify that $$\tag{1}$$

satisfies the $$\Omega$$-axiom.$${}^\dagger$$ The construction is defined in the first link above.

For convenience: here $$(A, f)\stackrel{k}{\rightarrowtail}(B, g)$$ is an arbitrary, monic $$\mathbf{Bn}(I)$$-arrow, taken as an inclusion; $$(I, \operatorname{id}_I)$$ is the terminal object in $$\mathbf{Bn}(I)$$; $$p_I$$ is the projection $$p_I(\langle x, y\rangle)=y$$; $$\top$$ is defined by $$\top(i)=\langle 1, i\rangle$$; and $$\chi_k$$ is the product map $$\langle\chi_A, g\rangle$$, i.e., $$\chi_k(x)=\begin{cases}\langle 1, g(x)\rangle &: x\in A \\ \langle 0, g(x)\rangle &: x\notin A.\end{cases}$$

Thoughts: What I've done so far is replace $$\chi_k$$ with an arbitrary $$\mathbf{Bn}(I)$$-arrow $$h: \langle B, g\rangle\to \langle 2\times I, p_I\rangle$$ in $$(1)$$, supposing what I get is a pullback. Then I've run it though the definition of a pullback quite easily. I've had numerous stupid ideas about what to do next (with all manner of confusing diagrams) but to no avail.

I'd like a detailed solution, please.

It should be easier than I think it is. Maybe my problem is with bundles themselves. This is my second attempt at reading Goldblatt's book: last time I thought I had'm but got up to "11.4: Models in a Topos" - right where I wanted to be - before other commitments made me lose track entirely; now I'm about to read "4.8: $$\Omega$$ and comprehension".

$$\dagger$$: The $$\Omega$$-axiom is given on page 81, ibid., via the definition of a subobject classifier:

Definition: If $$\mathbb{C}$$ is a category with a terminal object $$1$$, then a subobject classifier for $$\mathbb{C}$$ is a $$\mathbb{C}$$-object $$\Omega$$ with a $$\mathbb{C}$$-arrow $$\text{true}: 1\to\Omega$$ that satisfies the following axiom.

$$\Omega$$-axiom: For each monic $$f:a\rightarrowtail d$$ there is one and only one $$\mathbb{C}$$-arrow $$\chi_{f}:d\to\Omega$$ such that $$\chi_f\circ f=\text{true}\circ !$$ is a pullback square.

$\require{AMScd}$ First, recall that $\mathbf{Bn}(I)$ is just a notation for the slice category $\mathsf{Set}_{/I}$.

Lemma. For any category $\mathscr C$ and any object $c$ of $\mathscr C$, the forgetful functor $\mathscr C_{/c} \to \mathscr C$ commutes with fibered products.

So if you have a pullback as in your question, the square $$\begin{CD} A @>k>> B \\ @VfVV @VV\chi_k V \\ I @>>\top> 2 \times I \end{CD}$$ is a pullback in $\mathsf{Set}$. Then, remark that the square $$\begin{CD} I @>\top>> 2\times I \\ @VVV @VVV \\ 1 @>>\mathrm{true}> 2 \end{CD}$$ is also a pullback ($\mathrm{true}$ being the map selecting $1 \in 2$). So, concatenating the two squares makes the outer square of $$\begin{CD} A @>k>> B \\ @VfVV @VV\chi_k V \\ I @>>\top> 2 \times I \\ @VVV @VVV \\ 1 @>>\mathrm{true}> 2 \end{CD}$$ a pullback again. But then $2$, equipped with the map $\mathrm{true} \colon 1 \to 2$, is a subobject classifier for $\mathsf{Set}$. From here, you can easily derive the uniqueness of $\chi_k$ (remember that $p_I \circ \chi_k$ is fixed to be $g$ by hypothesis).

• Thank you, @Pece. I had to make sure I can understand this without recourse to functors, though, given the way Goldblatt has structured his book; I can just about see what you've done :) Jun 25, 2014 at 11:45

Just to fill in the gaps of the great answer above. Understand how PBL (pullback lemma) works in Goldblatt. And understand that for products there exists a unique arrow $$\langle p, q \rangle$$ for any third object $$a \xleftarrow{p} c \xrightarrow{q} b$$ onto the argument objects (of the product). Note that $$! \circ f = !$$ and that in set $$\chi_A$$ is already known to exist (regardless of uniqueness) since $$\textbf{Set}$$ has a subobject classifier $$\Omega$$. Thus after pasting the two pullbacks, you get a pullback diagram on the right (the square). But more importantly, we know that the morphism $$\chi_A$$ is the unique such morphism that creates that pullback square by the $$\Omega$$ axiom in $$\text{Set}$$.

Thus $$\chi_k = \langle m, g \rangle$$ where $$m = \chi_A$$. Now apply the product rule to the object $$B$$ with its two projectors onto the arguments of $$2 \times I$$, namely $$\chi_A : B \to 2$$ and $$g:B\to I$$. Then by definition of $$\langle \chi_A, g\rangle$$ it is the unique such arrow (we call it $$\chi_k$$) such that $$p_2 \circ \chi_k = \chi_A$$ and $$p_I \circ \chi_k = g$$. That's using the universal property of product.

But that is the same as saying it is the unique arrow such that the square on the top left (together with arrows into $$I$$) above is a pullback in $$\text{Bn}(I)$$ because one of the conditions that the whole thing commutes or in particular that $$p_I \circ \chi_k = g$$.

As is often the case in mathematics, we don't always make full use of a condition, i.e. we only said "such that $$p_I \circ \chi_k = g$$" and didn't mention all the other conditions going on in the pullback. One condition was enough in this case.

• It's a slight abuse of notation to write $!\circ f=!$, is it not? Apr 18, 2020 at 0:18
• @Shaun yes, that's correct. ! is used anytime there's a unique something and we don't need to name it. Apr 18, 2020 at 12:42