Can you get a long exact sequence from a short exact sequence of abelian groups together with another abelian group?
Solution
Let \[ 0\longrightarrow A'\longrightarrow A\longrightarrow A''\longrightarrow 0 \] be a short exact sequence of abelian groups, and let \(B\) be another abelian group.
1. Applying the left-exact functor \(\Hom(B,-)\) and its right derived functors gives a long exact sequence\[ 0\to\Hom(B,A')\to\Hom(B,A)\to\Hom(B,A'')\to\Ext^1(B,A')\to\Ext^1(B,A)\to\Ext^1(B,A'')\to\Ext^2(B,A')\to\cdots. \]
In the category of abelian groups, \(\Hom(B,-)\) is left exact. Its right derived functors are \(\Ext^i(B,-)\). The general long-exact-sequence theorem for right derived functors applied to the displayed short exact sequence gives the stated sequence.
2. Applying the contravariant left-exact functor \(\Hom(-,B)\) gives\[ 0\to\Hom(A'',B)\to\Hom(A,B)\to\Hom(A',B)\to\Ext^1(A'',B)\to\Ext^1(A,B)\to\Ext^1(A',B)\to\cdots. \]
The functor \(\Hom(-,B)\) is contravariant and left exact. Its right derived functors in the first variable are \(\Ext^i(-,B)\), and the variance reverses the arrows among the \(\Hom\) and \(\Ext\) terms as displayed.
3. Applying the right-exact functor \(-\otimes B\) and its left derived functors gives the Tor long exact sequence\[ \cdots\to\Tor_2(A'',B)\to\Tor_1(A',B)\to\Tor_1(A,B)\to\Tor_1(A'',B)\to A'\otimes B\to A\otimes B\to A''\otimes B\to0. \]
The functor \(-\otimes B\) is right exact, and its left derived functors are \(\Tor_i(-,B)\). The general long-exact-sequence theorem for left derived functors yields the displayed exact sequence.
Thus a short exact sequence of abelian groups, together with a third abelian group, naturally produces the standard long exact \(\Ext\) and \(\Tor\) sequences.