Let $X$ be a compact Riemann surface. Is the sheaf of meromorphic function $\mathcal M_X$ a flasque sheaf? Remember that flasque means that the restrition morphisms are surjective.


Nope. Consider, for example, the Riemann sphere; call it $X$. There are very few global (edit: bijective) meromorphic functions on $X$. The only ones are the fractional linear transformations, $$ z\mapsto\frac{az+b}{cz+d}. $$ But when you remove one point from $X$ it becomes $\mathbb C$, which has many more meromorphic functions.

Edit: When I wrote this, I was thinking of biholomorphic functions on the Riemann sphere. Of course, there are other meromorphic functions than the ones I gave, such as $z\mapsto z^2$. What I should have said is that every meromorphic function on $X$ is a rational function; and of course, there are meromorphic functions on $\mathbb C$ that are not the restriction of rational functions.

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    $\begingroup$ isn't $z \mapsto z^2$ meromorphic on the Riemann sphere ? $\endgroup$ – mercio Oct 17 '16 at 13:44
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    $\begingroup$ All the rational functions $z\mapsto p(z)/q(z)$ with $p,q$ polynomials are meromorphic on the Riemann sphere. Removing one point $z_0$ from it brings in functions with an essential singularity at $z_0$ though. And those are not restrictions of meromorphic functions. $\endgroup$ – Jyrki Lahtonen Oct 17 '16 at 13:48

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