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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 !

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    $\begingroup$ For the second question, there is a very easy answer with abstract nonsense: as a coproduct of presheaves is a colimit, it is preserved by any left-adjoint functor. In particular, sheafification, being left adjoint to the inclusion Sheaves $\to$ Presheaves, preserves a coproduct $\endgroup$ – Maxime Ramzi Jun 13 '18 at 18:29
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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)

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    $\begingroup$ nice. I need to learn more catergory theory, it helps make the arguments cleaner. My answer for question 2 was a bit clumsy. I am glad you confirmed, via a more "high-tech" point of view. $\endgroup$ – Malkoun Jun 13 '18 at 18:44
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    $\begingroup$ @Malkoun : once the right notions are in place, you can let category theory guide you gently ;-) (for question 2.; neither argument I'm using is complicated once you know the definitions of the words used) $\endgroup$ – Maxime Ramzi Jun 13 '18 at 18:48
  • $\begingroup$ I guess part of me likes to keep things very concrete and as explicit as possible, but this makes what I write quite difficult to read, and I get bogged down in detail. I do appreciate though the power of looking at maps rather than objects, well exploited by Grothendieck. I do appreciate the power of abstract nonsense, but I tend not to think that way. Do you have any good reference for category theory for geometers, with an emphasis on its connections with algebraic geometry (as in "Categories for the Working Geometer"?). $\endgroup$ – Malkoun Jun 13 '18 at 18:57
  • $\begingroup$ @Malkoun unfortunately I don't personnally know any such references, but looking around a bit on MO gives the following questions (and answers !) : mathoverflow.net/questions/216215/… and mathoverflow.net/questions/212548/… $\endgroup$ – Maxime Ramzi Jun 13 '18 at 19:24
  • $\begingroup$ Thanks a lot Max and Malkoun, I’ll try to learn the basics to understand your argument which seems to be almost straightforward as soon as you know the definitions and concepts. $\endgroup$ – BrianT Jun 13 '18 at 20:53
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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.

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  • $\begingroup$ Thanks a lot, I'll check the properties. Do you have an idea for the second question ? $\endgroup$ – BrianT Jun 13 '18 at 15:03
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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.

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  • $\begingroup$ Hi. I followed the link, but could not find the counterexample. Do you mean in Ravi Vakil's notes? $\endgroup$ – Malkoun Jun 14 '18 at 4:33
  • $\begingroup$ Hi Fan Zheng, so can you tell me where the above argument of Max fails : let $F^l$ be sheaves, and $F$ be the functor sheafification. Then $$F(\underset{l}{\bigoplus} F^l) = \underset{l}{\bigoplus} F(F^l) = \underset{l}{\bigoplus} F^l$$ $\endgroup$ – BrianT Jun 14 '18 at 6:29
  • $\begingroup$ I did some search, and indeed, some counterexamples can be found for example here: math.stackexchange.com/questions/193816/… $\endgroup$ – Malkoun Jun 14 '18 at 7:04
  • $\begingroup$ @BrianT Sheafification doesn't commute with direct sum as the definition of sheafification involves products, possibly infinite, and infinite products don't commute with infinite sums. $\endgroup$ – Will Sawin Jun 14 '18 at 7:37
  • $\begingroup$ @WillSawin probably the actual reason is that in my "proof" the second $\bigoplus$ is the direct sum in $\mathbf{Sh}(X)$ (I realized my blunder too late) and this does not coincide with the direct sum in $\mathbf{Psh}(X)$ (the inclusion functor does not commute with direct sums, while the sheafification functor does) $\endgroup$ – Maxime Ramzi Jun 14 '18 at 9:13

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