Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The category $\Gamma^{\mathrm{op}}$ is defined to be a skeleton of the category of finite pointed sets (see also this question). Then $\Gamma$-spaces, meaning space-valued presheaves $\Gamma^{\mathrm{op}}\to \mathrm{Spaces}$, can be used to present spaces with commutative and associative multiplication up to all higher homotopies. This is similar to how the category $\Delta^{\mathrm{op}}$ can be defined as a skeleton of the category of finite total orders with distinct endpoints, and presheaves $\Delta^{\mathrm{op}}\to \mathrm{Spaces}$ (simplicial spaces) can be used to present spaces with associative multiplication up to all higher homotopies.

It is known that the topos of simplicial sets is the classifying topos for total orders with distinct endpoints. Does the topos of set-valued presheaves on $\Gamma$ have a similar interpretation?

share|improve this question
4  
Isn't the category of presheaves on $fSet^{op}$ (the opposite category of finite sets) the topos that classifies objects (so geometric morphisms $E\to PSh(fSet^{op})$ correspond to objects of $E$)? If so, I'd guess that $PSh(fSet_*^{op})$ classifies based objects of $E$ ... –  Charles Rezk Jan 13 '12 at 21:05
add comment

2 Answers 2

up vote 7 down vote accepted

I agree with Charles Rezk's comment. Quite generally, the classifying topos for a universal Horn theory $T$ is the topos of covariant set-valued functors on the category of finitely presented models of $T$. This is proved (twice) in an old joint paper of mine and Andre Scedrov's, "Classifying topoi and finite forcing," J. Pure Appl. Algebra 28 (1983) 111-140. (For searching, it may help to know that the paper is so old that it uses the original form of Andre's name, "Andrej" and with \v accents on the S and c of his last name.)

Apply that description of the classifying topos to the theory with no axioms in a vocabulary consisting of just one constant symbol, i.e., the theory of pointed sets. Its finitely presented models are just the finite pointed sets, so its classifying topos is the topos of covariant functors on the category $\Gamma^{\text{op}}$ of the question. Equivalently, it's the topos of pre sheaves on $\Gamma$.

share|improve this answer
2  
In a sense, this example is easier than the case of $\Delta$, because linearity of an ordering is not a universal Horn statement. An analogous treatment of the $\Delta$ case would begin with the universal Horn theory of partial orders with distinct top and bottom, impose a Grothendieck topology on its classifying topos to force linearity of the order, and invoke Grothendieck's "Lemme de comparaison" to simplify the resulting sheaf topos to a presheaf topos on a sub-site that turns out to be $\Delta$. –  Andreas Blass Jan 16 '12 at 2:18
    
Thank you! I should have been able to figure that out for myself (I knew the fact about universal Horn theories), but I think I got confused about what is $\Gamma$ and what is $\Gamma^{op}$. –  Mike Shulman Jan 17 '12 at 5:32
add comment

Of course the above answer of @AndreasBlass exhausts it all, but it may be still amusing to compute directly the statement in the previous comment by @CharlesRezk.

One additional piece of information is this: for an object $X$ in a topos $T$, giving a geometric morphism from a topos $E$ to $T/X$ is equivalent to giving a geometric morphism $f:E\to T$ together with a point $1\to f^*(X)$.

Applying this to $T=Psh(fSet^{op})$ and $X=$ the generic object, which is the embedding $X:(fSet^{op})^{op}\to Set$ gives that the category of pointed objects of $E$ is equivalent to the category of gm's from $E$ to $Psh(fSet^{op})/X$. Now the latter is equivalent to presheaves over the Grothendieck construction of $X$ which in turn is equivalent to the category of finite pointed sets (objects are pairs $(n,i)$ where $n$ is a finite set and $i$ is an element of $X(n)$, i.~e. of $n$, etc....), as a particular case of the equivalence $$ Psh(C^{op})/\hom_C(c,-)\simeq Psh((c/C)^{op}) $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.