surjective geometric morphism


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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.


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. .


Surjection/embedding factorization


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.


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.


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

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


With observation 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. 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.


Trivially, any connected geometric morphism is surjective.


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.


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.