Covering sieves in a Grothendieck topology

Question

I'm trying to get my head around some of the basics of Grothendieck topologies.

Let $(\mathcal{C}, J)$ be a site, let $U$ be an object of $\mathcal{C}$ and let $J(U)$ be the set of covering sieves on $U$.The axioms for $J$ being a Grothendieck topology on $\mathcal{C}$ are:

1. The maximal sieve $M = \bigcup_{A\in\mathcal{C}} \mathcal{C}(A,U)$ is in $J(U)$;
2. If $R\in J(U)$ and $f: V\rightarrow U$ is a morphism of $\mathcal{C}$ then the pullback sieve $f^* (R)$ defined as $\left\{W\xrightarrow{\alpha} V: f\circ\alpha\in R\right\}$ is in $J(V)$;
3. If $R\in J(U)$ and $S$ is another sieve on $U$ such that for every morphism $f: V\rightarrow U$ in $R$ we have $f^* (S)\in J(V)$ then $S\in J(U)$.

Johnstone (p.13) claims that the following is a consequence of axioms 1 and 3: if $R\in J(U)$ and $S$ is a sieve on $U$ containing $R$ then $S\in J(U)$. I have tried to prove this: let $f: V\rightarrow U$ be a morphism in $R\subseteq S$. Then axiom 2 implies that $f^* (R)\in J(V)$ and $f^* (S)$ is a sieve on $V$ containing $f^* (R)$.

I'm probably missing something obvious, but since $S$ is not necessarily in $J(U)$ I can't just use axiom 2 to conclude $f^* (S)\in J(V)$ and then axiom 2 to conclude that $S\in J(U)$. And I don't really see how to use axiom 1 here. Can anyone help me out?

Answer 1

Recall that a Grothendieck sieve $R$ defined on a category $C$ is a subclass of the class of objects of $C$ such that if $X$ is in $R$ and $f:Y\rightarrow X$, then $Y$ is in $R$.

Suppose defined a Grothendieck topology on $C$, let $R$ be an element of $J(U)$ ($J(U)$ is a subclass of $C/U$ the category over $U$) and $S$ a sieve containing $R$, to show that $S$ is in $J(U)$ it is enough to show that for every $f:X\rightarrow U\in R$, $S^f\in J(X)$, but $S^f=\{g:Y\rightarrow X: f\circ g\in S\}$, but $g$ is a morphism between $f\circ g$ and $f$ $(f\circ g$ and $f$ are objects of $C/U$ here), thus since $R$ is a sieve, $g\circ f\in R$ thus $S^f=R^f\in J(X)$.