Homological algebra and sheaves

Question

Consider two exact diagram of sheaves of abelian groups on some topological space $X$ $$ 0\rightarrow \mathcal{A} \rightarrow \mathcal{B} \rightarrow \mathcal{C} \rightarrow 0$$ $$ 0\rightarrow \mathcal{D} \rightarrow \mathcal{E}\rightarrow \mathcal{F} \rightarrow 0$$ with with an epimorphism of complexes of abelian sheaves such that $\mathcal{A}(X)\rightarrow \mathcal{D}(X)$ and $\mathcal{B}(X)\rightarrow \mathcal{E}(X)$ are surjective. My question is if then $\mathcal{C}(X)\rightarrow \mathcal{F}(X)$ also is surjective?

Answer 1

I'll illustrate a way to approach questions like this using the spectral sequence associated to a double complex. Recall that a spectral sequence is a collection $\{E_r,d_r\}$ of abelian groups $E_r$ with maps $d_r:E_r\rightarrow E_r$ such that $d_r^2=0$, with isomorphisms $E_{r+1}\cong H(E_r)$, the cohomology of $E_r$ with respect to the differential $d_r$.

This is a lot of data, and I won't explain it how we obtain in this case, but this general process is explained in lots of places, for example, Eisenbud's Commutative algebra book.

In our case, $E_r$ will be graded, and the differential $d_r$ will have bidegree $(r,r-1)$, so we can view this is a grid of abelian groups with arrows, and we "take cohomology" to get access to the next "level" of arrows.

Our spectral sequence "converges" in the sense that for all $(a,b)$ eventually the differential coming into, and out of the $(a,b)$ spot is zero. So in our case, we are using the spectral sequence associated to the total complex of that diagram.

The general theorem we are using is that there is a spectral sequence with $E_0$ page just the double complex itself, with initial differentials the "down" arrows, and the $E_1$ page the cohomology with respect to these, with the $E_1$ differentials the "across" arrows induced on cohomology. Then there are the (more opaque) later pages and higher differentials. Now if your double complex had exact rows, by swapping the roles of across/down, we see that this different $E_1'$ page is zero, which tells us that the cohomology of the total complex is zero, so in our original spectral sequence, everything will need to cancel out eventually. This is a lot of generalities, and I'm not justifying these assertions really, but part of the charm is that you don't need to know how the inner parts work necessarily to use spectral sequences, like how you don't need to understand how an engine works to be able to drive.

So lets see what this looks like in your diagram. First, we take global sections, and the long exact sequence in cohomology to get the first diagram, I've drawn the relevant part only. I've also simplified the notation, and am using $A,B,C$ to mean the global sections of these sheaves. Then we have the zeroth, first, and second pages of the spectral sequence, using $A'$ to mean cohomology with respect to the previous differential.

We know that everything eventually cancels out, and this will occur for us on the third page, since then all differentials are zero. So the $d_2$ differentials are all isomorphisms. For instance, we see that $A'',B'',C'',H^1(A)''$ are all zero already on this page. Now to actually answer your question, note that if $H^1(D)'$ and $H^1(B)''$ are zero, then we have the differentials out of $F'$ and $F''$ are both trivial, so we have $F'=F''=0$. This then implies that your original map was surjective on the global sections of $F$. This is implied if both $B$ and $D$ are flasque, but one could use these less clear but more precise weaker conditions too. One may also note here that this is a totally general technique, and more generally proves all the basic homological lemmas very quickly, and tells you how to discover them yourself for more exotic diagrams.