I'm stuck on Exercise 4.5.1 of Goldblatt's, "Topoi: A Categorial Analysis of Logic".

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

enter image description here$\tag{1}$

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".

Please help :)

$\dagger$: 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.


2 Answers 2


$\require{AMScd}$ 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).

  • $\begingroup$ 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 :) $\endgroup$
    – Shaun
    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).

Bn(I) Omega axiom satisfied picture

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.

  • $\begingroup$ It's a slight abuse of notation to write $!\circ f=!$, is it not? $\endgroup$
    – Shaun
    Apr 18, 2020 at 0:18
  • 1
    $\begingroup$ @Shaun yes, that's correct. ! is used anytime there's a unique something and we don't need to name it. $\endgroup$ Apr 18, 2020 at 12:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .