A few general questions about pre-sheaves and sheaves

Question

I am no specialist in sheaf theory, so I would be glad to get some help regarding the following:

I have a pre-sheaf $F$ of abelian groups above a topological space $X$, and I have found an open cover $\{U_i\}$ of $X$ such that, for any $i$, $F(U_i)$ is a direct sum: $$F(U_i) = \underset{l}{\bigoplus} F^l (U_i),$$ where $F^l$ are pre-sheaves on $X$.

I have the two following questions:

1. Is it true in general that a direct sum of pre-sheaves (resp. sheaves) of abelian groups is a pre-sheaf (resp. sheaf), or does it have to be a finite sum ? If it is true, how do we define sections and restriction morphisms for general direct sums of pre-sheaves ?
2. How to prove that the sheafification $F^{\#}$ of $F$ is given by the direct sum: $$F^{\#} = \underset{l}{\bigoplus} F^{l \#},$$ where $F^{l \#}$ is the sheafification of $F^l$ ?

Thanks a lot for your help !

Answer 1

This is an "abstract nonsense" answer. If you're not familiar with a bit of category theory this will probably be useless.

For the second question :

Denote by $L$ the sheafification functor $\mathbf{Psh}(X) \to \mathbf{Sh}(X)$.

By definition (or construction,...) $L$ is left adjoint to the inclusion $i: \mathbf{Sh}(X) \to \mathbf{Psh}(X)$; hence it preserves colimits.

Now let $(F^l)_l$ be a family of presheaves, $\displaystyle\bigoplus_lF^l$ can be defined as in Malkoun's answer and is actually the coproduct of the $F^l$'s in $\mathbf{Psh}(X)$; hence $L(\displaystyle\bigoplus_lF^l) = \displaystyle\bigoplus_lL(F^l)$ which is exactly what you wanted.

Edit : there's a mistake in what comes right above : $L$ does preserve colimits, but the inclusion functor need not do so as well; hence the second $\bigoplus$ is not the direct sum in $\mathbf{Psh}(X)$, it is the direct sum in $\mathbf{Sh}(X)$; which can be defined using $L$. Hence the answer to your question is "yes if the second $\bigoplus$ is taken in $\mathbf{Sh}(X)$ and not in $\mathbf{Psh}(X)$.". In particular, taking sections over $U$ need not be the sum of the sections.

By the way you can answer the first question in the same way: $\mathbf{Ab}$ is an abelian cocomplete category, hence $\mathbf{Psh}(X)$ is also abelian cocomplete; and also for any open set $U\subset X$, the "evaluation at $U$" functor preserves colimits, hence $(\displaystyle\bigoplus_lF^l)(U)$ is $\displaystyle\bigoplus_lF^l(U)$. Restriction morphisms come from the universal property of the coproduct, as described in Makoun's answer ($\displaystyle\bigoplus_l$ can in fact be seen as a functor $\mathbf{Ab}^I \to \mathbf{Ab}$ where $I$ is the set of indices for this very reason)

Answer 2

The answer to 1 was too long as a comment, so I will write it here.

Regarding 1, the answer is 'yes', meaning that one can define the notion of a direct sum of presheaves of abelian groups, but please check the details.

Suppose we have presheaves $F^l$ of abelian groups on a topological space $X$. We want to define their direct sum

$F = \bigoplus_l F^l$

First, we define, for $U$ an open subset of $X$,

$F(U) = \bigoplus_l F^l(U)$ .

Moreover, if $V \subseteq U$ is an open subset of $X$ contained in $U$, we denote the restriction homomorphisms of $F^l$ by $h^l_{VU}$, so that

$h^l_{VU}: F^l(U) \to F^l(V)$ .

We now define the restriction homomorphism $h_{VU}: F(U) \to F(V)$ as follows. Let $i_k(W): F^k(W) \to \bigoplus_l F^l(W)$ be the natural homomorphisms, where $W$ is an arbitrary open subset of $X$.

Consider the homomorphisms $g^l_{VU} = i_l(V) \circ h^l_{VU}: F^l(U) \to F(V)$. By the universal property of coproducts, satisfied by direct sums of abelian groups, there is a unique homomorphism $h_{VU}: F(U) \to F(V)$ such that $g^l_{VU} = h_{VU} \circ i_l(U)$ for any $l$.

Thus we have defined restriction homomorphisms $h_{VU}$, and it remains to check that they satisfy the required properties.

Answer 3

I am just adding a tiny bit that is missed from the previous 2 answers, that is, the resp. part of Question 1:

If $F^l$ are sheaves on $X$, is $\oplus_l F^l$, which exists in the category of presheaves on $X$, automatically a sheaf on $X$?

The answer is NO. For a counterexample see this answer.