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Recall that the group $Ext^1(F'',F')$ parametrizes extensions $$0 \rightarrow F' \rightarrow F \rightarrow F'' \rightarrow 0$$ as follos: given one such extension, consider the long exact cohomology sequence arising from the functor $Hom(F'',\bullet)$. If $\delta$ is the connecting coboundary map $$\delta:Hom(F'',F'') \rightarrow Ext^1(F'',F')$$ and we set $\theta\in Ext^1(F'',F')$ to be the image of the identity map on $F''$ under $\delta$, mthis process gives a 1-1 correspondence between isomorphism classes of extensions of $F''$ by $F'$, and elements of the group $Ext^1(F'',F')$.

Note that if $E''$ is locally free, we have an isomorphism. $Ext^1(F'',F')=H^1(F''^{\ast}\otimes F')$. I have read that the coboundary map $\delta$ is actually obtained by taking cup-product with the extension class $\theta$.

I was wondering whether someone could provide some insight on this last statement.

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Am I right that your $E$'s and $F$'s are (quasi)coherent sheaves on a scheme and that $E''$ is actually $F''$? – Serge Lvovski Feb 1 '13 at 8:33
This follows from a general fact concerning the behaviour of coboundary maps on cup products, namely that $\delta(u\cup v) = \delta u \cup v$ under certain hypothesis. Yours is the case $u=1$. See Brown's "Cohomology of groups", V.3.3. – Mark Grant Feb 1 '13 at 8:52
@Mark: Isn't this an answer? – Martin Brandenburg Feb 1 '13 at 9:59
up vote 2 down vote accepted

This follows from a general fact concerning the behaviour of coboundary maps on cup products. Let $$ 0\to M'\to M\to M''\to 0 $$ be a short exact sequence, and let $N$ be an object such that the sequence $$ 0\to M'\otimes N \to M\otimes N \to M''\otimes N\to 0$$ is exact. Then $\delta(u\cup v)=\delta u\cup v$ for any $u\in H^p(X;M'')$ and $v\in H^q(X;N)$. Yours is the case $u=1$.

See Brown's "Cohomology of groups", V.3.3 for the case of modules over a group ring, or Bredon's "Sheaf theory", II.7.1(b) for the case of sheaf cohomology.

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I was trying to come up with a precise and detailed proof, but ended up in being stuck. If you take $u\in H^p(X;M'')$ to be $1$ (I guess that by $1$ you mean the neutral element of the cohomology group) you do not have something related to the exact sequence you started with, because $H^p(X;M)=\mathrm{Ext}^p(O_X,M'')\neq \mathrm{Ext}^p(M',M'')$... – Filippo Alberto Edoardo Jun 10 '15 at 15:24

The Yoneda product is the following:

$$Ext^n(M,N)\otimes Ext^m(L,M)\rightarrow Ext^{n+m}(L,N).$$

For $L=\mathbb{Z}$ and $n=0,1$ and M,N sheaves of $\mathbb{Z}$-modules this specializes to

$$Hom(M,N)\otimes H^m(M)\rightarrow H^{m}(N),$$ $$Ext^1(M,N)\otimes H^m(M)\rightarrow H^{m+1}(N).$$

Given a short exact sequence of sheaves of $\mathbb{Z}$-modules

\begin{equation} 0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0, \end{equation}

applying the functor $Hom(C,\cdot)$ yields a long exact sequence in which the first boundary map is

$$\delta:Hom(C,C)\rightarrow Ext^1(C,A)$$

and the extension class $\xi$ of the given short exact sequence is the image of the identity under $\delta$, i.e. $$\delta(1_C)=\xi$$ see e.g. Hartshorne Ch.III, Exc.6.1. We may now put the two above Yoneda products as rows in the following diagram

$$\begin{array}{ccc} Ext^1(C,A)\otimes H^m(C)&\rightarrow& H^{m+1}(A)\\ \uparrow\ \delta\otimes\operatorname{id}&&\uparrow\ \delta\\ Hom(C,C)\otimes H^m(C)&\rightarrow &H^{m}(C) \end{array}$$

Now Bredon's "Sheaf theory", II.7.1(b), (see the answer by Mark Grant) gives the commutativity of this diagram, i.e. $\delta(u\cup v)=\delta(u)\cup v$. Setting $u=1_C$ yields

$$\delta(v)=\xi\cup v$$ as desired.

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