isomorphism and surjection of sheaves of rings

Question

I think something of my understanding of morphism of sheaves of rings went wrong. Below it seems I made a self contradicting result.

Let $\mathscr F,\mathscr G$ be sheaves of rings defined on a topological space $X$. Let $\varphi: \mathscr F\rightarrow \mathscr G$ be a surjection of sheaves of rings and let $\mathscr I$ denote the kernel. Note that this does not mean $\varphi_{U}:\mathscr F(U)\rightarrow \mathscr G(U)$ is surjective for all $U$. But it means the induced map on stalks $\varphi _{x}:\mathscr F_x\rightarrow \mathscr G_x$ is surjective for all $x$. Then if I take the quotient, the map $(\mathscr F/\mathscr I)_x\rightarrow \mathscr G_x$ should be an isomorphism for all $x$. If it is an isomorphism on all stalks, then the morphism $\mathscr F/\mathscr I\rightarrow \mathscr G$ is an isomorphism for all $U\subset X$ open. But the map $\mathscr F/\mathscr I \rightarrow \mathscr G$ being an isomorphism on all $U$ means that $\mathscr F\rightarrow \mathscr G$ is a surjection on all $U$. This is kind of contradicting that $\varphi$ is not necessarily surjective for all $U\subset X$ open.

Where did I make a mistake understanding the isomorphism of sheaves?

Answer 1

The first issue here is that you didn't really do anything - all you've done is restate "if $F\to G$ is surjective, then why isn't $F(U)\to G(U)$ surjective for all $U$" as "if $F\to F/I$ is surjective, then why isn't $F(U)\to (F/I)(U)$ surjective for all $U$" by replacing $G$ with $F/I$.

The specific claim you make here that is wrong is the following:

But the map $\mathscr F/\mathscr I \rightarrow \mathscr G$ being an isomorphism on all $U$, means $\mathscr F\rightarrow \mathscr G$ is surjection on all $U$...

The issue here is that $G\cong F/I$ may have a section over $U$ which is not a section of $F$, potentially because we removed an obstruction that prevents a local section from extending to a global section. Let's look at an example with $U=X=\Bbb A^2\setminus 0$, $F=\mathcal{O}_X$, $G=i_*\mathcal{O}_Y$ where $i:Y\to X$ is the inclusion of the punctured $Y$-axis. Then $F\to G$ is certainly surjective since $i$ is a closed immersion, but the induced map on global sections is not surjective: it's $k[x,y]\to k[y^{\pm 1}]$ by $x\mapsto 0$, $y\mapsto y$. The section $y^{-1}$ comes from the section $y^{-1}$ on the complement of the $x$-axis, which cannot extend to a global section of $\mathcal{O}_X$.

Answer 2

You never make any use of the multiplication so this is a question about sheaves of abelian groups. The best way to understand how an argument has gone wrong is to check it on an example. Let's check the best motivating example of a nontrivial short exact sequence of sheaves, the exponential sheaf sequence

$$1 \to 2 \pi i \mathbb{Z} \to \mathcal{O}_X \xrightarrow{\exp} \mathcal{O}_X^{\times} \to 1$$

on $X = \mathbb{C}^{\times}$. Here $F = \mathcal{O}_X$ is the sheaf of holomorphic functions, $G = \mathcal{O}_X^{\times}$ is the sheaf of invertible holomorphic functions, and $\varphi : F \to G$ is the exponential, with kernel the constant sheaf $I = 2 \pi i \mathbb{Z}$.

So, $\varphi$ is a surjection of sheaves. This amounts to the statement that invertible holomorphic functions locally have logarithms, which is true. The map on stalks $F_x \to G_x$ is also surjective, which is the same statement for germs of invertible holomorphic functions. And since $G \cong F/I$ it holds that $(F/I)_x \cong G_x$. Here is where the mistake is:

But the map $\mathscr F/\mathscr I \rightarrow \mathscr G$ being an isomorphism on all $U$ means that $\mathscr F\rightarrow \mathscr G$ is a surjection on all $U$.

In our example you are now claiming that $F(U) \to G(U)$ is always a surjection, which is of course false; in particular, famously $F(X) \to G(X)$ is not surjective, because the identity function $\mathbb{C}^{\times} \to \mathbb{C}^{\times}$ does not have a global logarithm. The issue, as KReiser says, is exactly that a section of $F/I \cong G$ does not necessarily lift to a section of $F$ - indeed if this were true then no sheaf would have nontrivial cohomology and there would be no need to study sheaf theory!

More abstractly, the issue is exactly that the section functor $F \mapsto F(U)$ is not necessarily exact, so it does not necessarily preserve epimorphisms, and we do not necessarily have $(F/I)(U) \cong F(U)/I(U)$. This is why the section functor has interesting derived functors!