# Why can a subpresheaf of a representable functor be seen as a set of arrows closed under precomposition?

A subpresheaf $$Y$$ of $$X$$ is a presheaf $$Y$$ such that for all $$C$$ $$YC \subseteq XC$$ and $$Yf$$ is $$Xf$$ restricted to $$Y(cod(f))$$. In my lecture notes is the following quote:

"What do subpresheaves of $$\hom(-, C)$$ look like? If $$R$$ is a subpresheaf of $$\hom(-, C)$$, then $$R$$ can be seen as a set of arrows with codomain $$C$$ such that if $$f: C'\to C$$ is in $$R$$, and $$g: C'' \to C'$$ is arbritrary, then $$fg$$ is in $$R$$ (for $$fg = \hom(f, C)(g)$$). "

I don't understand the last part.

As you said, a subpresheaf of $$\operatorname{hom}(-, C)$$ is a presheaf $$Y$$ such that for each $$C'$$ we have $$YC' \subseteq \operatorname{hom}(C', C)$$. So $$Y$$ is just a collection of arrows with codomain $$C$$. But it is not just any such collection, because for $$g: C'' \to C'$$ we must have that $$Yg$$ is the restriction of $$\operatorname{hom}(g, C)$$. The latter is just given by precomposition: it takes some $$f: C' \to C$$ in $$\operatorname{hom}(C', C)$$ and outputs $$fg: C'' \to C$$ in $$\operatorname{hom}(C'', C)$$. So it should be $$Yg(f) = fg = \operatorname{hom}(g, C)(f)$$ and not what you wrote in your question. But then $$fg \in YC''$$, so since $$g$$ was arbitrary we have that $$Y$$ is closed under precomposition.
So a subpresheaf $$Y$$ of $$\operatorname{hom}(-, C)$$ gives us a collection of arrows into $$C$$ that is closed under precomposition. Conversely, given a collection $$R$$ of arrows into $$C$$ we can define a subpresheaf $$Y$$ of $$\operatorname{hom}(-, C)$$ by setting $$YC' = \{f \in R : \operatorname{dom}(f) = C'\}$$ and then $$Yg$$ is of course the restriction of $$\operatorname{hom}(g, C)$$ for any $$g: C'' \to C'$$. The fact that this is well-defined is due to the fact that $$R$$ was assumed to be closed under precomposition.