# Biproduct of sheaves is a sheaf

Let $C$ be an additive category. By that I mean, the composition map is bilinear, the existence of a zero object and the existence of biproducts, where biproduct means the following:
An object $X_1 \oplus X_2$ is said to be a biproduct of objects $X_1$ and $X_2$ if there exist inclusion maps $i_1, i_2$ and projection maps $p_1, p_2$ such that $p_k \circ i_l=id_{X_k}$, if $k=l$, and $0$ else, and $i_1 \circ p_1 + i_2 \circ p_2 = id_{X_1 \oplus X_2}$.

Now let $S_1, S_2$ be $C$-valued sheaves on some fixed topological space $T$ and suppose that the objects in $C$ are (at least) sets. There came up a problem when proving that $S_1 \oplus S_2$ given by $(S_1 \oplus S_2)(U):=S_1(U) \oplus S_2(U)$ as again a sheaf. It is no problem to prove that it is again a presheaf. Let $p_{S_1(U)}$ and $p_{S_2(U)}$ denote the projections from $S_1(U) \oplus S_2(U)$ to $S_1(U)$ and $S_2(U)$, respectively.

Suppose $s,t$ are elements in $S_1(U) \oplus S_2(U)$ such that they locally coincide (with respect to some open cover $(U_i)$ of $U$). Using the sheaf property of $S_1$ and $S_2$, I proved that $p_{S_i(U)}(s)=p_{S_i(U)}(t)$ for $i=1,2$. However, I don't seem to be able to get anything else out of it since I am not sure if in general an element in $S_1(U) \oplus S_2(U)$ is of the form $(p_{S_1(U)}(s),p_{S_2(U)}(s))$, i.e. is uniquely determined by its projections. Using the identity $i_1 \circ p_1 + i_2 \circ p_2 = id_{X_1 \oplus X_2}$ doesn't help either since I cannot assume the objects of $C$ to have any algebraic structure, which is in addition respected by the morphisms.
The biproduct defined above is in particular a product (with the projections). Since the cartesian product of sets is such a product, I may assume that elements in $S_1(U) \oplus S_2(U)$ are of the form $(p_{S_1(U)}(s),p_{S_2(U)}(s))$, noting that the product is unique up to unique isomorphism. That is the only thing I can think of to move one.

To not make this post any longer, I just note that I will probably run into trouble again when checking the second sheaf axiom.

Any help is highly appreciated.

• You are talking about «objects in $S_1(U)$» but since $S_1(U)$ is an object of your abstract category $C$, that makes no sense. – Mariano Suárez-Álvarez Nov 9 '15 at 17:13
• You are right, so let us assume that objects in $C$ are sets and then talk of elements in those sets. – Plankton Nov 9 '15 at 17:14
• Morevoer, an additive category does have enough structure to define a canonical addition in hom-sets with respect to which composition is bilinear. This should be explained in any textbook treating additive categories! (Like MacLane's) – Mariano Suárez-Álvarez Nov 9 '15 at 17:14
• Well, you not only need to be able to talk about elements of objects of $C$, but are assuming that the direct sum in $C$ is the cartesian product of the underlying sets or something like that and so on. – Mariano Suárez-Álvarez Nov 9 '15 at 17:16
• Thought the definition of the bilinearity made clear that the hom-sets are supposed to be abelian groups. – Plankton Nov 9 '15 at 17:16

## 1 Answer

Here is a proof that the pointwise product of two $C$-valued sheaves is a sheaf, for any category $C$ with binary products (the argument generalizes to any limits, actually). Let $S_1$ and $S_2$ be $C$-valued sheaves and let $S$ be the presheaf defined by $S(U)=S_1(U)\times S_2(U)$, with the obvious restriction maps. Let $\{U_i\}$ be a collection of open sets with $U=\bigcup U_i$. To show that $S$ is a sheaf, we must show that the restriction maps make $S(U)$ the limit of the diagram consisting of the objects $S(U_i)$ and $S(U_i\cap U_j)$ and the restriction maps $S(U_i)\to S(U_i\cap U_j)\leftarrow S(U_j)$ for each $i$ and $j$. So suppose you have an object $Q$ of $C$ together with maps $Q\to S(U_i)$ such that for each $i$ and $j$, the compositions $Q\to S(U_i)\to S(U_i\cap U_j)$ and $Q\to S(U_j)\to S(U_i\cap U_j)$ are equal. Since $S(U_i)=S_1(U_i)\times S_2(U_i)$, composing our maps with the two projections yields maps $Q\to S_1(U_i)$ such that the compositions $Q\to S_1(U_i)\to S_1(U_i\cap U_j)$ and $Q\to S_1(U_i)\to S_1(U_i\cap U_j)$ are equal, and similarly for $S_2$. Since $S_1$ and $S_2$ are sheaves, these maps determine maps $Q\to S_1(U)$ and $Q\to S_2(U)$, which can be combined to a map $Q\to S(U)$ which makes the required diagram commute. This map is also unique, since a map $Q\to S(U)$ is determined uniquely by its projections to $S_1(U)$ and $S_2(U)$, and the maps $Q\to S_1(U)$ and $Q\to S_2(U)$ were the unique maps that made the required diagram commute.

If you find this proof hard to parse, it may be helpful to remark that it is almost exactly the same as a typical proof that the cartesian product of two sheaves of sets is a sheaf. The only difference is that every time you would refer to an element of $S(U)$ for some $U$, you instead refer to a map $Q\to S(U)$.

• Thanks for the proof, I understand it. (You may want to change "$Q \to S_1(U_i) \to S_1(U_i \cap V_i)$" to $Q \to S_1(U_j) \to S_1(U_i \cap U_j)$). Now I know that the product of two sheaves (with values in a category $C$ with binary products) is again a sheaf. That it is a biproduct of $X_1$ and $X_2$ just follows from $X_1(U) \times X_2(U) = X_1(U) \oplus X_2(U)$ being a biproduct of $X_1(U)$ and $X_2(U)$ in $C$, right? – Plankton Nov 10 '15 at 11:49
• Right, once you know it's a sheaf, you can just construct the maps making it a biproduct separately on each open set, and then check that they are compatible with the restrictions. – Eric Wofsey Nov 10 '15 at 18:38