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surjective geometric morphism

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Definition

Definition

A geometric morphism between toposes (f *f *):(f^* \dashv f_*) : \mathcal{E} \to \mathcal{F} is surjective or a geometric surjection if it satisfies the following equivalent criteria:

The equivalence of these condition appears for instance as MacLaneMoerdijk, VII 4. lemma 3 and prop. 4.

Proof

We discuss the equivalence of these conditions:

The equivalence (f *faithful)(Idf *f *ismono)(f^* \; faithful) \Leftrightarrow (Id \to f^* f_* \;is \; mono) is a general property of adjoint functors (see there).

The implication (f *faithful)(f *inducesinjectiononsubobjects)(f^* faithful) \Rightarrow (f^* induces\;injection\;on\;subobjects) works as follows:

first of all f *f^* does indeed preserves subobjects: since it respects pullbacks and since monomorphisms are characterized as those morphisms whose domain is stable under pullback along themselves.

To see that f *f^* induces an injective function on subobjects let UXU \hookrightarrow X be a subobject with characteristic morphism charU:XΩchar U : X \to \Omega and consider the image

f *U f *** f *X f *charU f *Ω. \array{ f^* U &\to& f^* * \simeq * \\ \downarrow && \downarrow \\ f^* X &\stackrel{f^* char U}{\to}& f^* \Omega } \,.

of the pullback diagram that exhibits UU as a subobject. Since f *f^* preserves pullbacks, this is still a pullback diagram.

If now UU˜U \leq \tilde U but f *(U)=f *(U˜)f^* (U) = f^*(\tilde U) then both corresponding pullback diagrams are sent by f *f^* to the same such diagram. By faithfulness this implies that also

U˜ * X charU Ω \array{ \tilde U &\to& * \\ \downarrow && \downarrow \\ X &\stackrel{char U}{\to}& \Omega }

commutes, and hence that also U˜U\tilde U \subset U, so that in fact U˜U\tilde U \simeq U.

Next we consider the implication (f *inducesinjectiononsubobjects)(f *isconservative)(f^* induces injection on subobjects) \Rightarrow (f^* is conservative).

Assume f *(XϕX)f^* (X \stackrel{\phi}{\to} X') is an isomorphism. We have to show that then ϕ\phi is an isomorphism. Consider the image factorization Xim(ϕ)XX \to im(\phi) \hookrightarrow X'. Since f *f^* preserves pushouts and pullbacks, it preserves epis and monos and so takes this to the image factorization

f *Xf *(imϕ)f *X f^* X \to f^* (im \phi) \stackrel{\simeq}{\to} f^* X'

of f *ϕf^* \phi, where now the second morphism is an iso, because f *ϕf^* \phi is assumed to be an iso. By the assumption that f *f^* is injective on subobjects it follows that also imϕXim \phi \simeq X' and thus that ϕ\phi is an epimorphism.

It remains to show that ϕ\phi is also a monomorphism. For that it is sufficient to show that in the pullback square

X× XX X ϕ X ϕ X \array{ X \times_{X'} X &\to& X \\ \downarrow && \downarrow^{\mathrlap{\phi}} \\ X &\stackrel{\phi}{\to}& X' }

we have X× XXXX \times_{X'} X \simeq X. Write Δ:XX× XX\Delta : X \to X \times_{X'} X for the diagonal and let

XimΔ ϕX× XX X \to im \Delta_\phi \to X \times_{X'} X

be its image factorization. Doing the same for f *ϕf^* \phi, which we have seen is a monomorphism, and using that f *f^* preserves the pullback, we get

f *imΔ ϕf *(X× XX). f^* im \Delta_\phi \simeq f^* (X \times_{X'} X) \,.

Now using again the assumption that f *f^* is injective on subobjects, this implies imΔ ϕ=X× XXim \Delta_\phi = X \times_{X'} X and hence that ϕ\phi is a monomorphism.

(…)

The statement about the comonadic adjunction we discuss below as prop. 2.

Properties

Surjection/embedding factorization

Observation

For T:T : \mathcal{E} \to \mathcal{E} a left exact comonad the cofree algebra functor

F:TCoAlg() F : \mathcal{E} \to T CoAlg(\mathcal{E})

to the topos of coalgebras is a geometric surjection.

Proof

By the discussion at topos of coalgebras the inverse image is the forgetful functor to the underlying \mathcal{E}-objects. This is clearly a faithful functor.

Proposition

Up to equivalence, every geometric surjection is of this form.

This appears for instance as (MacLaneMoerdijk, VII 4., prop 4).

Proof

With observation 1 we only need to show that if f:f : \mathcal{E} \to \mathcal{F} is surjective, then there is TT such that

f F TCoAlg(). \array{ \mathcal{E} &\stackrel{f}{\to}& \mathcal{F} \\ & {}_{\mathllap{F}}\searrow & \downarrow^{\mathrlap{\simeq}} \\ && T CoAlg(\mathcal{E}) } \,.

For this, take T:=f *f *T := f^* f_*. This is a left exact functor by definition of geometric morphism. By assumption on ff and using the equivalent definition of def. 1 we have that f *f^* is a conservative functor. This means that the conditions of the monadicity theorem are met, so so f *f^* is a comonadic functor.

For more on this see geometric surjection/embedding factorization . Also at monadic descent.

Examples

Trivially, any connected geometric morphism is surjective.

Proposition

For f:XYf : X \to Y a continuous function between topological spaces and (f *f *):Sh(X)Sh(Y)(f^* \dashv f_*) : Sh(X) \to Sh(Y) the corresponding geometric morphisms of sheaf toposes, ff is a surjection precisely if (f *f *)(f^* \dashv f_*) is a surjective geometric morphism.

References

Section VII. 4. of

Last revised on March 14, 2018 at 12:33:01. See the history of this page for a list of all contributions to it.