Composing covers with epis

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 $Y$?
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2022-07-25 20:47:06
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Answers: 1

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
2022-07-25 22:24:38
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