1
$\begingroup$

I am beginner of sheaf-theory and beg your pardon for this maybe silly question.

Let $\mathcal{C}$ be a Grothendieck site and $T$ the category of sheaves on $\mathcal{C}$ and let $f:X\rightarrow Y$ be an epic morphism in $T$ into a representable sheaf $Y$.

I have a general lack of understanding how such epic morphisms look like and this leads to the the questions:

  • Suppose $\{Y_\alpha\to Y\}$ is a cover of $Y$. Is $Y_\alpha\times_Y X$ representable? I see no formal reason why this is true but somehow my miserable intuition still thinks it might be.
  • (If the answer to the first point is ''no'', please assume a representable $X$) Suppose $\{X_\alpha\to X\}$ is a cover of $X$. Is the composition $\{X_\alpha\to X\xrightarrow{f} Y\}$ a cover of $Y$?
  • 0
    The fibre product of two representable sheaves over a representable sheaf is itself representable... if the site itself has those fibre products.2012-05-16

1 Answers 1

0

In more detail.

  1. By definition, limits commute with hom: $$\mathcal{C}(-, \varprojlim x_\bullet) \cong \varprojlim \mathcal{C} (-, x_\bullet)$$ In particular, fibre products commute with hom: $$\mathcal{C}(-, x \times_z y) \cong \mathcal{C}(-, x) \times_{\mathcal{C}(-, z)} \mathcal{C}(-, y)$$ So far this has nothing to do with Grothendieck topologies. If you have a subcanonical Grothendieck topology then all this still applies verbatim; otherwise you have to replace $\mathcal{C}(-, x)$ with its sheafification, and the claim only holds for finite limits because sheafification is only left exact.

    Now, if $Y$ and $Z$ are representable but $X$ is not, then it does not follow that $X \times_Z Y$ is representable. For example, $\mathcal{C}$ could be a site with finitely many objects and finitely many morphisms, while $X$ (and hence $X \times_Z Y$ could be infinite.)

  2. I presume by ‘cover’ you mean jointly epimorphic. The answer is yes. In a Grothendieck topos, all small coproducts exist, so you can replace a jointly epimorphic family $\{ X_\alpha \to X \}$ with a single epimorphism $\coprod_\alpha X_\alpha \to X$. It is a fact about general categories that the latter morphism, when it exists, is epimorphic if and only if the original family is jointly epimorphic. The composition of two epimorphisms is obviously an epimorphism, hence, $\{ X_\alpha \to X \to Y \}$ is jointly epimorphic if $X \to Y$ is an epimorphism.

    The question is slightly more subtle if we work in the underlying site. There, one cannot take coproducts (even if they exist) and one must look at the Grothendieck topology $J$ itself. One of the axioms of Grothendieck topologies is that covers compose: if $\mathfrak{U}$ is a sieve on $Y$ such that the pullback sive $f^* \mathfrak{U}$ is a $J$-covering sieve on $U$ for each $f : U \to Y$ in $\mathfrak{U}$, then $\mathfrak{U}$ is also a $J$-covering sieve. In terms of a Grothendieck pretopology $K$, if $\{ U_\beta \to Y \}$ is a $K$-covering family and $\{ X_{\alpha, \beta} \to U_\beta \}$ is a $K$-covering family for each $U_\beta$, then the composite family $\{ X_{\alpha, \beta} \to U_\beta \to Y \}$ is also a $K$-covering family.

  • 0
    I need help with your second answer. In my situation $f:X\rightarrow Y$ is a morphism of the site $\mathcal{C}$ such that the *Yoneda embedded* $\tilde f:\tilde X\rightarrow\tilde Y$ is en epi of sheaves. If now $\{X_\alpha\rightarrow X\}$ is a sieve on $X$, is the composition $\{X_\alpha\rightarrow X\rightarrow Y\}$ a sieve on $Y$? My problem is then the relation between epis (or jointly epimorphic families) in the site $\mathcal{C}$ and epis (or jointly epimorphic families) in the category of sheaves on $\mathcal{C}$. How are they related?2012-05-18
  • 0
    A family of morphisms generates a $J$-covering sieve in the site if and only if it becomes jointly epimorphic in the sheaf topos. ($J$ really has to be a Grothendieck topology for this to be true.)2012-05-18
  • 0
    Wow, this would be perfect. I think, I can show that if $\{X_\alpha\rightarrow X\}$ is a cover, then it becomes jointly epimorphic in the sheaf topos. Do you know how I can show the converse?2012-05-18
  • 0
    The joint _presheaf_ image of $\{ \tilde{X}_\alpha \to \tilde{X} \}$ is a _presheaf_ subobject of $\tilde{X}$, and there is a bijective correspondence between _presheaf_ subobjects of $\tilde{X}$ and (not necessarily covering) sieves on $X$. If the joint _sheaf_ image is the whole of $\tilde{X}$, then the joint _presheaf_ image is dense, and a dense _presheaf_ subobject of $\tilde{X}$ is exactly a $J$-covering sieve.2012-05-18
  • 0
    I think I understand it better now, thanks. Doesn't this mean, that when I take a Grothendieck pretopology $K$ which is maximal in the sense that it can't be refined anymore without changing the associated Grothendieck topos, then a family $\{\tilde X_\alpha\rightarrow \tilde X\}$ of representables is jointly epimorphic iff $\{X_\alpha\rightarrow X\}$ is a cover of $K$?2012-05-21
  • 0
    Grothendieck pretopologies are still relatively well behaved. If $K$ is a Grothendieck pretopology and $J$ is the Grothendieck topology it generates, then a family is $K$-covering if and only if the sieve it generates is $J$-covering.2012-05-21
  • 0
    Suppose $F=\{X_\alpha\to X\}$ is a family of morphisms of the site with Grothendieck pretopology $K$ and generated Grothendieck topology $J$. You are saying that "$F$ becomes jointly epimorphic in the sheaf topos" $\Leftrightarrow$ "$F$ generates a $J$-covering sieve" $\Leftrightarrow$ "$F$ is a $K$-covering", right? But there are different Grothendieck pretopologies $K$, $K'$ that give rise to the same Grothendieck topology $J$ and then the sheaf topos is the same and certainly has the same epimorphisms. Doesn't one have to require some saturation condition on $K$?2012-05-22
  • 0
    @Shinze: Sorry, I misspoke. A sieve is $J$-covering if and only if it contains _some_ $K$-covering family. So you have to look at the sieve a family generates.2012-05-22
  • 0
    Sorry, but I am confused. "$F$ becomes jointly epimorphic in the sheaf topos" $\Leftrightarrow$ "$F$ generates a $J$-covering sieve" is still correct, right? I would like to see that this implies that $F$ is a $K$-covering in a *maximal* Grothendieck pretopology $K$ giving rise to $J$. This $K$ is defined as $F\in K(X)\Leftrightarrow \{f\circ g\mid f\in F\}\in J(X)$. It should be implied that for such a maximal Grothendieck pretopology $K$ and a family $F$: "$F$ becomes jointly epimorphic in the sheaf topos" $\Leftrightarrow$ "$F$ is a $K$-covering", right?2012-05-22
  • 0
    If this is true, it would answer the second point of my initial question together with your answer. By *cover* in the initial question, I meant $K$-covering where $K$ is a *maximal* Grothendieck pretopology in the above sense.2012-05-22
  • 0
    The maximal Grothendieck pretopology $\overline{K}$ generating a Grothendieck topology $J$ can be described explicitly as follows: a family is $\overline{K}$-covering if and only if it generates a $J$-covering sieve. So the short answer is yes.2012-05-22