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I am taking my first Algebraic Geometry class and I do not understand how to solve the following exercises.

1) Let $Q$ be the sheaf of sections of the map $f:S^1 \rightarrow S^1$ defined via complex coordinates as $f(z)=z^2$, i.e. $$ Q(\mathcal{U}):=\{s:\mathcal{U}\rightarrow S^1 | f \circ s(z)=z \}. $$ Show that $Q$ has no global sections.

2) Now let $Q_k$ be the sheaf which assigns to each open set $\mathcal{U}$ the $k$ vector space freely generated by the set $Q(\mathcal{U})$. By taking a carefully chosen cover of $S^1$, show that $$ F: \mathcal{U} \rightarrow Q_k(\mathcal{U}) \otimes Q_k(\mathcal{U}) $$ is not a sheaf.

Can anyone help me?

Thank you

Edit. 1) As @conditionalMethod suggested, I solved the first problem as follows. Assume that a global section $\tilde{s}$ exists. Then $\tilde{s}$ is a continuous function defined from $S^1$ to $S^1$ and it is such that $f(\tilde{s}(z))=z$. So $\deg(f)\cdot \deg(\tilde{s})=1$, but since $\deg(f)=2$ then we have $\deg(\tilde{s})\notin \mathbb{Z}$, which is a contradiction.

2) For the second problem my guessing is that, when tensoring the elements via pointwise multiplication, there will be some cross-multiple terms that will prevent the agreement on the intersections, but I do not understand how to choose the right cover.

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  • $\begingroup$ What have you tried? Where, specifically, do you get stuck? $\endgroup$
    – KReiser
    Commented Nov 27, 2019 at 20:51
  • $\begingroup$ 1) For the first I tried to mimic another proof, but I'm not quite sure of what's happening: assume by contradiction that a global section $\tilde{s}$ exists; thus $\deg(\tilde{s})=0$, but if I consider $g(z)=z^2\tilde{s}(z)$, then $g$ has degree $0$, so $\tilde{s}$ has degree $-2$ $\Rightarrow$ contradiction. $\endgroup$
    – mat95
    Commented Nov 27, 2019 at 21:00
  • $\begingroup$ 2) I have no idea of how to "carefully choose" the cover. $\endgroup$
    – mat95
    Commented Nov 27, 2019 at 21:01
  • $\begingroup$ (1) as written is false, but very likely it is missing a condition on the elements of $Q$, like continuity or stronger. With that assumption, a global section is an element of $Q(S^1)$. This is, a (continuous) function $s:S^1\to S^1$ such that $f(s(z))=z$. Then $\deg(f)\deg(g)=1$, but $\deg(f)=2$, which would leave $\deg(g)$ being not an integer. $\endgroup$ Commented Nov 27, 2019 at 21:04
  • $\begingroup$ @mat93 you should add your attempts to the body of your post via the edit button. Putting what you've tried in your post will help you get better answers. $\endgroup$
    – KReiser
    Commented Nov 27, 2019 at 21:06

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