# isomorphism between category of sheaves and morphisms of abelian groups

I am working on theory of category and I found this exercise. I tried a lot but I didn't know how I could do. Let $$A$$ a discrete valuation ring. Show that the category of sheaves of abelian groups on $$Spec(A)$$ is equivalent to the category which objects are defined as below:

$$\{ f: S \rightarrow L \ \quad S,L\in Ab \}$$ and if we take two morphisms $$f: S \rightarrow L$$ and $$f': S' \rightarrow L'$$ then $$g\circ f = f' \circ g'$$ with $$g: L \rightarrow L'$$ and $$g': S \rightarrow L$$.

• This is almost immediate from the definitions. Have you tried writing down the definitions of everything involved? Do you know what $\operatorname{Spec}(A)$ is in this case? – Eric Wofsey Oct 8 '18 at 18:38
• OK, so, what are the open sets of $\operatorname{Spec}(A)$? What data does a sheaf consist of? – Eric Wofsey Oct 8 '18 at 18:46

Let us write $$X=\operatorname{Spec} A$$. It seems that your main confusion is about what the topology on $$X$$ looks like in this case. If $$A$$ is a discrete valuation ring, it has two prime ideals $$P=\{0\}$$ and $$Q$$, the maximal ideal. So $$X=\{P,Q\}$$.
Now we need to determine the topology on $$X$$. By definition, a subset of $$X$$ is open iff it is a union of sets of the form $$D(f)=\{x\in X:f\not\in x\}$$ for elements $$f\in A$$. So, we must determine these sets $$D(f)$$. If $$f=0$$, then $$D(f)=\emptyset$$. If $$f\not\in Q$$, then $$f$$ is a unit, so $$D(f)=X$$. Finally, if $$f\in Q$$ is nonzero, then $$f\in Q$$ but $$f\not\in P$$, so $$D(f)=\{P\}$$. Since any union of these sets will again just give another one of these sets, we conclude that there are three open subsets of $$X$$: $$\emptyset,\{P\}$$, and $$X$$.
So a presheaf $$F$$ on $$X$$ consists of three abelian groups $$F(X)$$, $$F(\{P\})$$, and $$F(\emptyset)$$ together with restriction homomorphisms $$F(X)\to F(\{P\})$$ and $$F(\{P\})\to F(\emptyset)$$. For $$F$$ to be a sheaf, it needs to satisfy the gluing axiom, but in this case it is rather trivial, since no open subset of $$X$$ can be written as a union of other open subsets in a nontrivial way. The only restriction imposed is that $$F(\emptyset)$$ must be a trivial group, since $$\emptyset$$ is covered by the union of no open sets (this is true for a sheaf on any space).
So, since $$F(\emptyset)$$ must be trivial, we lose no information by ignoring it, and the data we are left with is two abelian groups $$F(X)$$ and $$F(\{P\})$$ together with a homomorphism $$F(X)\to F(\{P\})$$. This is exactly the data of the category you are asked to show is equivalent. I will leave it to you to write out all the details and verify that this really does give an equivalence of categories.
• Yes, you want to show the functor taking $F$ to the restriction map $F(X)\to F(\{P\})$ is an equivalence of categories. (I had a typo before where I wrote $F(\{Q\})$ instead of $F(\{P\})$, now fixed.) I don't know what diagram you are talking about being commutative. – Eric Wofsey Oct 8 '18 at 19:42
• Our sheaf is a sheaf on $X$, not a sheaf on a one point space. They're totally different things, even if our space happens to have a subset that has one pont... – Eric Wofsey Oct 9 '18 at 0:18