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I was looking for an example of two sheaves that satisfy the gluing and uniqueness axioms but whose tensor product is a presheaf, but not a sheaf (doesn't satisfy uniqueness, for example).

Thank you very much.

HallaSurvivor
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Werner Germán Busch
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    So to clarify, by "sheaf" you mean a sheaf of abelian groups, and by "tensor product" you mean the pointwise tensor product of presheaves of abelian groups? – Qiaochu Yuan Oct 19 '15 at 23:20
  • By Sheaf I mean a presheaf of $A$-Modules with $A$ a principal ideal domain that satisfies the gluing and uniqueness condition. So it's slightly more specific. – Werner Germán Busch Oct 19 '15 at 23:45
  • And yes, by tensor product I mean: the pointwise tensor product of the associated $A$ modules. – Werner Germán Busch Oct 19 '15 at 23:46

1 Answers1

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Consider the sheaves $\mathcal{F}=\mathcal{G}=\underline{\mathbb{Z}}$ of locally constant functions with values in $\mathbb{Z}$. Almost by definition : $\mathcal{F}(U)=\mathbb{Z}^{\pi_0(U)}$.

Now consider an open subset $U$ with two connected components. You have $$\mathcal{F}(U)\otimes\mathcal{G}(U)=\mathbb{Z}^{\oplus 2}\otimes\mathbb{Z}^{\oplus 2}=\mathbb{Z}^{\oplus 4}.$$

However $\mathcal{P}:V\mapsto\mathcal{F}(V)\otimes\mathcal{G}(V)$ is not a sheaf (in fact its associated sheaf is again $\underline{\mathbb{Z}}$). To see this, consider the open covering of $U$ by its two components $U_1$ and $U_2$. If $\mathcal{P}$ was a sheaf, we would have the equality $\mathcal{P}(U)=\mathcal{P}(U_1)\oplus\mathcal{P}(U_2)$. But it doesn't hold.

Roland
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  • YAY THANKS! you man – Werner Germán Busch Oct 21 '15 at 03:58
  • "we would have the equality P(U)=P(U1)⊕P(U2)P(U)=P(U1)⊕P(U2)." Why is that? Sheafs separate different connected components but presheafs do not? – Werner Germán Busch Oct 21 '15 at 04:01
  • @WernerGermánBusch The constant presheaf, or the tensor presheaf above, does not separate connected components. In fact, why should they ? There is no reason for this in the definition of presheaf. However sheaves satisfies the equality $F(U)=F(U_1)\oplus F(U_2)$ if $U_1$ and $U_2$ are the connected components of $U$. This follows immediately from the sheaf condition with the covering of $U$ by $U_1$ and $U_2$. – Roland Oct 21 '15 at 07:22
  • "The constant presheaf, or the tensor presheaf above, does not separate connected components. In fact, why should they ?"

    That's what I said.... lol

     – Werner Germán Busch Oct 26 '15 at 01:12
  • Why it doesn't hold? $F(U_1)=\mathbb{Z} \otimes \mathbb{Z}=\mathbb{Z^{\oplus 2}}$, isn't it? – jpatrick Feb 18 '18 at 00:17
  • @jpatrick No $\mathbb{Z}\otimes\mathbb{Z}=\mathbb{Z}$, the map $a\otimes b\rightarrow ab$ is an isomorphism. – Roland Feb 18 '18 at 00:21
  • I deleted my last question, because it's trivial and this isomorphism is also. Sorry for stupid questions... – jpatrick Feb 18 '18 at 00:42