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Stream: learning: questions

Topic: property of flasque presheaves?

Naso (Sep 21 2022 at 03:34):

Let $F : \mathcal{T}^{op} \to \mathsf{Set}$ be a presheaf on a topological space $(X,\mathcal{T})$ that is flasque, meaning whenever $i : U \hookrightarrow V$ the restriction map $F i : FV \to FU$ is surjective.

Then is the following true: for all $U,V$ and $x \in F U, y \in F V$ such that ${x \mid_{U \cap V}} = {y \mid_{U \cap V}}$, there exists $z \in F(U \cup V)$ with ${z \mid_U = x}$ and ${z \mid_V = y}$?

If not, is there a simple counterexample?

Naso (Sep 21 2022 at 04:02):

Ok, I think this may be a counterexample...

Let $X = \{a,b,c\}$ be a space generated by the two open sets $U = \{a,b\}, V = \{b, c\}$, so $\mathcal{T} = \{ \emptyset, \{b\}, U, V, X \}$.

Let $F$ be defined as follows:
$F \emptyset = \{ () \}$,
$F \{b\} := \{ (b_0) \}$,
$F U := \{ (a_0, b_0) , (a_1, b_0) \}$,
$F V := \{ (b_0, c_0), (b_0, c_1) \}$,
$FX:= \{ (a_0, b_0, c_1) , (a_1, b_0, c_0) \}$.

And the action of $F$ on inclusions is by restriction of tuples.

Then take $x = (a_0, b_0)$, $y = (b_0, c_0)$. Then these agree on restriction to $\{b\}$, but we cannot find $z \in F(U \cup V)$ restricting to both $x$ and $y$. Is this correct?

Morgan Rogers (he/him) (Sep 21 2022 at 08:19):

Seems right to me.

David Michael Roberts (Mar 15 2024 at 05:02):

I would even go so far as to say $FU = \{a_0,a_1\} \times \{b_0\}$, $FV = \{b_0\}\times \{c_0,c_1\}$, and $FX = \{b_0\}\times \{(a_0,c_1),(a_1,c_0)\}.$