# Trying to understand the equivalence of two definitions of a sieve.

Let $\mathcal{C}$ be a small category, let $C$ be an object of $\mathcal{C}$ and let $\mathbf{y}:\mathcal{C}\to[\mathcal{C}^{op},\mathbf{Set}]$ be the Yoneda embedding.

I am trying to derive the simple fact that a sieve $S$ on $C$ is a family of morphisms in $\mathcal{C}$, all with common codomain $C$, such that $f\in S\Rightarrow f\circ g\in S$ (whenever the composition makes sense) from the fact that $S$ is a subobject $S\subseteq \mathbf{y}(C)=\mathrm{Hom}_{\mathcal{C}}(-,C)$ but there must be something I do not get.

$S\subseteq \mathbf{y}(C)$ really means a monic $S\to\mathbf{y}(C)$ in the presheaf category. Hence it is a natural transformation between the functors $S$ and $\mathbf{y}(C)$. Therefore it is a collection of set functions $$\lbrace t_A:SA\to \mathbf{y}(C)(A)\rbrace_{A\in \textrm{ob}\mathcal{C}}$$ such that for all $g:B\to A$ in $\mathcal{C}$ the following square is commutative $$\require{AMScd} \begin{CD} SA @>{t_A}>> \mathrm{Hom}_{\mathcal{C}}(A,C);\\ @VVV @VVV \\ SB @>{t_B}>> \mathrm{Hom}_{\mathcal{C}}(B,C); \end{CD}$$ where the vertical arrows are $Sf$ and $-\circ g$.

But I don't see from here how the first description of a sieve above follows...I am probably not looking at this correctly. Can someone help?

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Actually, here it is best to use subpresheaf in the strictest possible sense: each $S A$ is a subset of $\mathcal{C}(A, C)$ and each map $S A \to \mathcal{C}(A, C)$ is the inclusion. –  Zhen Lin Nov 12 '13 at 8:45
Makes much more sense now! If I am not mistaken, you are saying that if you have a monic natural transformation then the family of maps it induces consists of monics as well? Is this a general fact? –  user121314 Nov 12 '13 at 9:58
That is true when the components are in a category where kernel pairs exist. –  Zhen Lin Nov 12 '13 at 13:57
...Like $\mathbf{Set}$... –  user121314 Nov 12 '13 at 14:00

The monic $S \rightarrow y(C)$ determines subsets $S(A) \subseteq Hom_\mathcal{C}(A,C)$, in the standard sense (why?), so that composing with a morphism sends the subset $S(A)$ to $S(B)$. Since the category is small, you can also take the union $\bigcup_{A \in \mathcal{C}} S(A)$. You can then check that this union is precisely a sieve, and that an isomorphic subobject will give the same sieve.
And of course, given an bunch of "equivariant subsets" as above, you can construct a subobject of $y(C)$.
I guess my problem was your "(why?)". It is because a monic natural transformation $S\to \mathrm{Hom}_{\mathcal{C}}(-,C)$ consists of monic components, right? For some reason this was not obvious for me, now it is. Thanks very much! –  user121314 Nov 12 '13 at 14:04