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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.