(This is from Chapter I.2 of Kollar's "Rational curves on algebraic varieties".)
Let $B$ be an extension of $A$: $0\rightarrow J\rightarrow B\rightarrow A\rightarrow 0$, $A$ and $B$ both being Noetherian local rings with residue field $k$. Let $T$ be a $k$-algebra, $I \subset T$ and ideal, $Q=T/I$, and $$R\xrightarrow r G\xrightarrow g I\rightarrow 0 \tag{1}$$ a presentation of $I$ by free $T$-modules.
Let $T^B$ be Noetherian, flat over $B$ such that $T^B\otimes_B k \cong T$, and the $m_B$-adic topology on $T^B$ is separated. Let $T^B\otimes_B A \cong T^A$. Let $I^A \subset T^A$ be an ideal such that $Q^A = T^A/I^A$ is flat over $A$.
We can lift (1) to an exact sequence of $T^A$ modules $$R^A\xrightarrow {r^A} G^A\xrightarrow {g^A} I^A\rightarrow 0. \tag{2}$$
Let $r^B$ and $g^B$ be any morphisms lifting $r^A$ and $g^A$ respectively. Then we have a commutative diagram $\require{AMScd}$ \begin{CD} R^B @>{r^B}>> G^B @>{g^B}>> I^B @>>> 0 &\text{(not necessarily a complex!)}\\ @VVV @VVV @VVV\\ R^A @>{r^A}>> G^A @>{g^A}>> I^A @>>> 0 &\text{(exact row)}\\ \end{CD}
We can then define $e(g^B):R\rightarrow Q\otimes J$ independently on the choice of $r^B$, and we can write $$E(g^B) = \text{coker}[R\xrightarrow {e(g^B)+r} Q\otimes J + G]$$ as an extension $$0\rightarrow Q\otimes J\rightarrow E(g^B) \rightarrow I \rightarrow 0.$$
It is said that we now obtain an element $E^B(Q^A) := [E(g^B)]\in \text{Ext}_T^1(I,Q\otimes J)$, which is called the obstruction.
Now finally here are my questions:
- How do we obtain this element?
- What does it stand for, intuitively? What does it "obstruct"?