Extension class and cup product Recall that the group $Ext^1(F'',F')$ parametrizes extensions $$0 \rightarrow F' \rightarrow F \rightarrow F'' \rightarrow 0$$ as follows: 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.
 A: 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.
A: 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.
