# Induced sequence of global sections

I'm reading Differential Analysis on Complex Manifolds by Raymond O. Wells. It states the following in the beginning of section 3 of chapter 2 on page 51:

Consider a short exact sequence of sheaves:

$$0 \to \mathcal{A} \to \mathcal{B} \to \mathcal{C} \to 0$$

Then it is easy to verify that the induced sequence

$$0 \to \mathcal{A}(X) \to \mathcal{B}(X) \to \mathcal{C}(X) \to 0$$

is exact at $\mathcal{A}(X)$ and $\mathcal{B}(X)$ but not necessarily at $\mathcal{C}(X)$. After that they give an example of failing exactness at $\mathcal{C}(X)$.

My problem is that Ii dont see why the induced sequence has to be exact at $\mathcal{A}(X)$ and $\mathcal{B}(X)$. I tried to find an answer in other books. In the book Sheaf Theory by Bredon they make the same statement in proposition 2.2 but they also state it is easy to verify.

I tried to come up with an argument of my own and got something like this (the argument is wrong i think): If we have the zero element $0 \in \mathcal{B}(X)$ this should induce a section $\tilde{0}$ of the stalks of $\mathcal{B}$ and should take the value of the zero element in every stalk. Then since the sequence of sheaves is exact (and thus exact in stalk level) we get that around every $x \in X$ there should be an open $V$ such that $\tilde{0}|_{V}$ is the image of the zero section in $\mathcal{A}(V)$ and since $\mathcal{A}$ is a sheaf we can patch these together to get a global zero section.

Also while i get the example that shows inexactness at $\mathcal{C}(X)$ i dont get why this should generally be the case.

Any help would be much appreciated

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Your argument shows that the $0$ section in $\mathcal B(X)$ is the image of the $0$ section in $\mathcal A(X)$, but that's far from establishing exactness of the sequence at either $\mathcal A(X)$ or $\mathcal B(X)$. Exactness at $\mathcal A(X)$ means that the only section in $\mathcal A(X)$ that maps to the $0$ section in $\mathcal B(X)$ is $0$, and this can be established by a stalk-wise argument similar to what you wrote (but starting with an element of $\mathcal A(X)$).
As for exactness at $\mathcal B(X)$, the fact that the composite map, from $\mathcal A(X)$ to $\mathcal C(X)$, is zero is again easy to check stalk-wise. The crux of the issue is that any section $s$ in $\mathcal B(X)$ that maps to zero in $\mathcal C(X)$ comes from a section in $\mathcal A(X)$. A stalk-wise argument gives you that $s$ can be lifted locally near each point $x\in X$; that is, $x$ has a neighborhood $U_x$ over which there is a section $q_x$ in $\mathcal A(U_x)$ that maps to (the restriction to $U_x$ of) $s$. You need to patch together all these $q_x$'s to a single section over all of $X$, which will then map to $s$. The necessary patching is possible, because $\mathcal A$ is a sheaf, provided the various $q_x$'s agree on the intersections of their domains. To see that $q_x$ and $q_y$ have the same restriction to $U_x\cap U_y$, use the fact that the difference of these restrictions maps to $0$ in $\mathcal B(U_x\cap U_y)$, and invoke the exactness at $\mathcal A(U_x\cap U_y)$.
The last part of your question seems to presuppose that the book claimed that the sequence is never exact at $\mathcal C(X)$. But there is no such claim; you quoted it as saying "not necessarily [exact] at $\mathcal C(X)$. – Andreas Blass Mar 2 '13 at 16:04
Thanks a lot for your explanation. I managed to write down a proper argument. What i meant with the last part (the part that you commented after your answer) was that i dont see why i cant use the same type of argument to show exatness at $\mathcal{C}(X)$. That is, i know the examples proof that sometimes the sequence is not exact at $\mathcal{C}(X)$ but where does the type of argument used to prove exactness at $\mathcal{B}(X)$ fail. – piertje Mar 2 '13 at 18:01
If you try to use the same argument to prove exactness at $\mathcal C$, you'll find yourself needing to patch together sections of $\mathcal B$ (just as I had to patch sections of $\mathcal A$). On the overlaps of neighborhoods like $U_x$ and $U_y$, the sections you want to patch will differ by sections that come from $\mathcal A$ (not from $0$ as in my answer). Unfortunately, you can't patch them unless they actually agree on the overlaps, so this argument fails. (Analyzing more closely the role of $\mathcal A$ in this failure, one can arrive at cohomology of sheaves.) – Andreas Blass Mar 4 '13 at 2:06