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In any presheaf topos, there exists an object called Lawvere's segment, which can be described as the presheaf $L:A^{op}\to Set$ such that for each object $a\in A$, $L(a)=\{x\hookrightarrow\ h_a: x\in Ob([A^{op},Set])\}$.

That is, $L$ assigns to each object $a\in Ob(A)$ the set of monomorphisms with target $h_a(\cdot):=Hom(\cdot,a)$
in $[A^{op},Set]$. Since all objects in $[A^{op},Set]$ can be given as colimits of the representable functors $h_a$ for $a\in A$, we can actually give a characterization of the functor represented by $L$ as the "sub-object assigning functor".

Lawvere's segment is useful, because it naturally has the structure of a cylinder functor (given by taking the cartesian product with a presheaf $X$). In fact, it is a theorem of Cisinski (very closely related to the specialization Jeff Smith's theorem to presheaf toposes) that every accessible localizer on a presheaf category admits the structure of a closed model category generated by the homotopical data (donnée homotopique is the term used by Cisinski, so this is my rough translation) of some set (as opposed to proper class) of arrows $S\subset Arr([A^{op},Set])$, some cellular model $\mathcal{M}$, and the canonical cylinder given by Lawvere's segment.

Then my question: How can we describe Lawvere's segment in $Set_\Delta$ explicitly, that is to say, geometrically?

Edit: A cellular model M on a presheaf topos is a set of monomorphisms $M$ such that $llp(rlp(M))$ is the class of all monomorphisms. Note that $llp$ and $rlp$ give the class of arrows with the left lifting property (resp. right lifting property) to the class of arrows given in the argument.

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4 Answers 4

up vote 4 down vote accepted

I've never heard this called 'Lawvere's segment' before, but your $L$ is the subobject classifier in the presheaf topos $[A^{\mathrm{op}},\mathrm{Set}]$. In presheaf toposes generally, the subobject classifier $\Omega$ is the presheaf that sends an object $a$ to the set of all sieves on $a$, i.e. the set of subobjects of $\hom(-,a)$. This means in this case that simplicial subsets $S' \subset S$ are in bijection with simplicial maps $S \to L$.

Off the top of my head, I can't be exactly sure what $L$ looks like, but I'd guess it's the constant simplicial set with $L_n = 2$ (the set of truth values) and all maps identities. Then the characteristic map $\chi_{S'} \colon S \to L$ will be the usual $s \mapsto 1$ if $s \in S'$ and $0$ otherwise, while the simplicial subset corresponding to some $\phi \colon S \to L$ will be given by the fibres $\phi_n^{-1}(1)$ in each degree. $S'$'s being a simplicial subset should correspond to $\phi$'s being a simplicial map.

I could easily be wrong about that last paragraph, though. There might be more about this in Mac Lane--Moerdijk.

Edit: I was wrong -- see the comments below.

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I'm sure I've seen it written 'Lawvere interval [object]', but a search brings up nothing. –  David Roberts Jun 18 '10 at 1:35
    
The description of $L$ is wrong. It cannot be $1 + 1$, or else simplicial sets would be a boolean topos. –  Andrej Bauer Mar 13 '11 at 11:04
    
@Andrej: I had accepted it for the observation that it is the subobject classifier. –  Harry Gindi Mar 13 '11 at 17:05
    
Yes, I understand that, I am just pointing out the answer has a mistake, lest someone gets the wrong idea. –  Andrej Bauer Mar 13 '11 at 17:20
    
Well, I did say I could easily be wrong :). Torsten's answer gives a clearer picture: $L_n$ is the set of sieves on [n] in $\Delta$. It may well be possible to give a more concrete description of these sets, given how $\Delta$ is generated from [0], [1] and [2]. –  Finn Lawler Mar 13 '11 at 17:39

The value of $L$ on the element $[n]\in\Delta$ is, by Yoneda's lemma, equal to the set of morphisms $[n]\rightarrow L$, i.e., the set of subobjects of $[n]$. Such a subobject $S$ is generated by the non-degenerate simplices of $S$ (as for a degenerate simplex $s$, the simplex of which it is a degeneration can be obtained by applying a simplicial operator to $s$) and hence corresponds to a simplicial complex with vertices in $[n]$. Conversely, any such simplicial complex gives rise to a subobject of $[n]$. Finally, if $f\colon[m]\rightarrow[n]$ is a morphism in $\Delta$, then $L([n])\rightarrow L([m])$ is given by the pullback of subobjects which translates into $S\mapsto f^{-1}(S)$.

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Torsten is right: an $n$-simplex of the simplicial set $L$ is a subcomplex of the standard $n$-simplex.

I don't know what to say about $L$ "geometrically", but here are a couple of observations: It is fibrant. (Every horn can be filled, usually in only one way, never more than three.) It is contractible.

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Edit: It appears that this is wrong! See the comments below.

Using Finn's observation that it is the subobject classifier, we can see that it is the nerve of the contractible groupoid with exactly two objects, $G_2$.

To see this, notice that $L_0=\{0,1\}$ induced by the two subobjects of $\Delta^0$, namely the empty map and the identity map. It has four 1-cells induced by the empty map, the inclusion of the first vertex, the inclusion of the second vertex, and the identity into $\Delta^1$. It has eight 2-cells, which we see by looking at the set of subobjects of $\Delta^2$ etc.

The nerve of $G_2$ has two $0$-cells, four $1$-cells, eight $2$-cells, etc.

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1  
I don't think this is reasonnable: the functor which takes a set $S$ to the contractible groupoid with $S$ as a set of objects is fully faithful, and so is the nerve functor. Therefore, you can count the number of maps $G_2\to G_2$: there are four of them. But you can find countably many maps $N(G_2)\to L$ (i.e. subobjects of the nerve of $G_2$): for instance, consider the skeletons $Sk^n(N(G_2))$, $n\geq 0$. Therefore $N(G_2)$ cannot be isomorphic to $L$. –  Denis-Charles Cisinski Mar 13 '11 at 21:47
    
Thanks for correcting my mistake! –  Harry Gindi Mar 13 '11 at 22:00

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