TITLE: Trying to understand the equivalence of two definitions of a sieve.
QUESTION [7 upvotes]: 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?

REPLY [3 votes]: 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 $y(C)$.