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enter image description here I have two questions about this exercise.

For (a), clearly if the topological space is irreducible, any open set is connected, so the sheaf (say a constant sheaf $\mathscr{A}$) is $A$ on each open set, and the restriction maps are isomorphisms. But does this also work if the topological space is not irreducible? $\mathscr{A}(U)$ would be a direct product of copies of $A$, one for each connected component. So the restriction map would be a map from a direct product of some copies of $A$ to a product of less copies of $A$, which is surjective?

For (c), why isn't it enough to have $\mathscr{F}$ is flasque? Say you have open sets $V \subset U$. Given any element $t'' \in \mathscr{F}''(V)$, by (b) there is a $t \in \mathscr{F}(V)$ that maps to $t''$; since $\mathscr{F}$ is flasque $t$ comes from some $s \in \mathscr{F}(U)$; let $s''$ be the image of $s$ in $\mathscr{F''}(U)$, then $s$ should restrict to $t$ by commutativity.

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  • $\begingroup$ Careful about connected compontents: $\mathscr{A}(U)$ only is the product over the components if they are open, which need not be true: If you take $\mathbb{Q} \subset \mathbb{R}$ with the subspace topology, every point is a connected component, because you can split every interval $(a, b) \subset \mathbb{Q}$ at an irrational point in between, so $(a, b) \cap \mathbb{Q} = ((a, c) \cup (c, b))\cap \mathbb{Q}$ for some $c \in \mathbb{R} \setminus \mathbb{Q}$ with $a < c < b$. $\endgroup$ – red_trumpet Apr 7 at 17:32
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In point (a), it is not true that if $U \subseteq V$ then it contains less connected components.

My original example does not work with the Zariski topolgogy, as pointed our in the comments, exactly because $\mathbb{A}^1$ is irreducible with that topology. Incidentally, it works with the Euclidean topology: $\mathbb{R}$ is an open connected subset of itself, but for example the complement of $\{0\}$ has two connected components. Hence the projection $\mathscr{A}(\mathbb{R}) \to \mathscr{A}(\mathbb{R} \setminus \{0\})$ boils down to $A \to A^2$. This is not necessarily surjective.

For a more algebraic example, consider the union of two affine lines, for example $X = V(xy)$ (where we are working in the ambient space $\mathbb{A}^2$), which is connected but not irreducible since $X = V(x) \cup V(y)$. The origin is a closed point, being equal to $V(x,y)$, so its complement is an open subset. However, it can be written as the union of two disjoint open subsets: they are given by the intersections with $V(x)$ and $V(y)$.

In (c), the requirement of $\mathscr{F}'$ flasque is necessary to invoke point (b). Otherwise, the exact sequence $$ 0 \to \mathscr{F}'(U) \to \mathscr{F}(U) \to \mathscr{F}''(U) $$ is only left exact, and your argument fails.

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  • $\begingroup$ I'm a bit confused: the closed points of $\mathbb{A}^1$ have the cofinite topology, so isn't the complement of $V(t)$ still connected? Or do you mean the real line with the Euclidean topology? $\endgroup$ – Takumi Murayama Dec 31 '17 at 17:14
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Suppose that $U$ and $V$ are disjoint nonempty open subsets of the connected space $X$. Consider the constant sheaf $\scr F$ with stalk $\Bbb Z$. Then ${\scr F}(U\cup V)$ has an element that is $1$ on $U$ and $0$ on $V$. This is not a restriction on any element of ${\scr F}(X)$ as these elements are all constant functions.

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