most recent 30 from http://mathoverflow.net 2013-05-21T19:37:13Z http://mathoverflow.net/feeds/question/120499 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120499/extension-class-and-cup-product Marc 2013-02-01T07:31:40Z 2013-02-01T12:41:23Z

<p>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 $Ext^1(F'',F')$.</p> <p>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 $\theta$.</p> <p>I was wondering whether someone could provide some insight on this last statement.</p>

http://mathoverflow.net/questions/120499/extension-class-and-cup-product/120513#120513 Mark Grant 2013-02-01T12:41:23Z 2013-02-01T12:41:23Z

<p>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 $u=1$. </p> <p>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.</p>