Let $X$ be a topological space, $\mathcal{F}$ a sheaf on $X$ and $K=\coprod_{x\in X}\mathcal{F}_x$.

Then let $\mathcal{G}$ be the sheaf of functions from $X$ to $K$. We can define a map of sheaves $i:\mathcal{F}\to\mathcal{G}\hspace{0.1cm}$ by letting $i_U:\mathcal{F}(U)\to\mathcal{G}(U)$ be given by $i_U:a\mapsto(f:x\mapsto a_x)$, where $a_x\in\mathcal{F}_x$ is the germ of $a$ at $x$ (this is compatible with restrictions since restriction doesn't change the germs).

Furthermore, it is injective: if $i_U(a)=i_U(b)$ then $a_x=b_x$ for all $x\in U$. Then $a$ and $b$ agree on a neighbourhood around every point of $U$, and so by the uniqueness part of the gluing axiom for sheaves we have $a=b$ in $\mathcal{F}(U)$.

This shows that we can regard any sheaf as a subsheaf of a sheaf of functions. Now let $\mathcal{F}$ be a presheaf, not necessarily a sheaf.

I believe that $i$ is still a map of presheaves. However, showing injectivity relied on the uniqueness part of the gluing axiom for sheaves. Then it doesn't seem like we can view $\mathcal{F}$ as a subpresheaf of $\mathcal{G}$ via $i$.

My question then is, can we view any presheaf as a subpresheaf of a presheaf of functions?

My first thought was to map $\mathcal{F}$ into its sheafification $\mathcal{F}^+$, and then include this into the associated sheaf of functions. But the answers to this question show that the inclusion to the sheafification is not necessarily injective.

Any help would be much appreciated.

  • $\begingroup$ In the given example, $K$ should be the product of the $\mathcal{F}_x$ instead of the coproduct. $\endgroup$ – withoutfeather Jul 30 '19 at 2:57
  • $\begingroup$ Maybe my notation isn't clear but I'm just taking $K$ to be the disjoint union of the stalks as sets so that I can map the germs to them, so I don't think we need to use the (Cartesian) product since we're never mapping to more than one stalk at a time? $\endgroup$ – Dave Jul 30 '19 at 3:04
  • $\begingroup$ Ah yes, you are right. $\endgroup$ – withoutfeather Jul 30 '19 at 3:59

The answer is no.

Indeed, every presheaf of functions you said earlier, i.e. a presheaf $\mathcal{G}$ such that $\mathcal{G}(U)$ is the set of functions $U\to K$, have to be a sheaf. But a presheaf map injectively to a sheaf implies that the presheaf maps injectively to its sheafification. Not every presheaf have this property (but this is true if your presheaf is separated).

To be clear, let $\mathcal{F}$ be a presheaf, $\mathcal{G}$ be a sheaf, and $\varphi:\mathcal{F}\to \mathcal{G}$ be an injective morphism of sheaves. Let $h:\mathcal{F}\to \mathcal{F}^{\#}$ be the canonical morphism from $\mathcal{F}$ to its sheafification, then $\varphi=\varphi^{\#}\circ h$. Since $\varphi$ is injective, it follows that $h$ must be injective.

Conversely, if a presheaf is separated, i.e. it maps injectively to its sheafification, then we can regard it as a subpresheaf of its sheafification. Now the sheafification is a subsheaf of a sheaf of functions, it implies that the presheaf is a subpresheaf of a sheaf of functions.

So my conclusion is: a presheaf can be regarded as a subpresheaf of a sheaf of functions if and only if it is separated.

Edit: If you still want to define a presheaf of functions, then you can take functions of a special case (not functions of sets or continuous functions, those functions can be glued and make your presheaf a sheaf).

  • $\begingroup$ Ah thanks I think I understand now, it's been helpful to see it's equivalent to being separated. I can picture now that when embedding the presheaf into a sheaf of functions, whilst we might have all of the functions in the target (because we can't glue), we certainly need those that do exist to agree if they are the same wherever they are defined. Thanks again! $\endgroup$ – Dave Jul 30 '19 at 4:19
  • $\begingroup$ *might not have all of the functions $\endgroup$ – Dave Jul 30 '19 at 4:36

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