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Let E be a Grothendieck topos, such as the category of sheaves of sets on a topological space. Then there is a unique geometric morphism $(\Delta \dashv \Gamma)\colon E\to \mathrm{Set}$, where $\Delta\colon \mathrm{Set}\to E$ constructs constant sheaves and its right adjoint Γ takes global sections. If E is locally connected (i.e. Δ has a further left adjoint), then Δ is a cartesian closed functor, i.e. $\Delta(B^A)\cong \Delta B^{\Delta A}$ for sets A,B.

Now in any cartesian closed category, we can define the "object of isomorphisms" Iso(X,Y) between any objects X,Y, as an equalizer of a pair of maps $X^Y \times Y^X \rightrightarrows X^X \times Y^Y$. In particular, when X=Y, we have the object Aut(X) of automorphisms of X. Since inverse image functors preserve finite limits, if E is locally connected then $\Delta(\operatorname{Aut}(X))\cong \operatorname{Aut}(\Delta X)$ for any set X.

My question is twofold:

  1. Is there a noticeably weaker condition on E than local connectedness which ensures that Δ preserves objects of automorphisms?

  2. Can you give an explicit example of a topos for which Δ does not preserve objects of automorphisms?

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Here is an answer to the second question. Consider the topos of sheaves on the one-point compactification of $\mathbb{N}$. Then for any set A, a global section of the constant sheaf ΔA is a function $\mathbb{N}\to A$ which is eventually constant. In particular, a global section of ΔAut(X) is an eventually constant sequence of automorphisms of X.

On the other hand, since ΔX is also the coproduct of X copies of the terminal object, a map ΔX → ΔX is just an X-indexed family of eventually constant functions $f_x\colon \mathbb{N}\to X$. In particular, an automorphism of ΔX is such a family such that for every n, the function $x \mapsto f_x(n)$ is an automorphism of X. So the difference between a global section of ΔAut(X) and an automorphism of ΔX is that the former is an eventually constant sequence of automorphisms, whereas the latter is a sequence of automorphisms such that for each $x\in X$, the sequence of images of x is eventually constant. Evidently the latter is strictly more general if X is infinite, so we cannot have $\Delta Aut(X) \cong Aut(\Delta X)$.

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