# 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 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

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 $u=1$.
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