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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(*). 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 :)


(*) 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.

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1 Answer 1

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$\require{AMScd}$ 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).

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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 :) –  Shaun Jun 25 at 11:45

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