Let $P_1\overset{\partial}{\rightarrow} P_0\rightarrow M\rightarrow 0$ be an exact sequence of $A$-modules with $P_0$, $P_1$ finitely generated and projective. The transpose $T(M)$ is defined as $\mbox{coker}\mbox{Hom}(\partial,A)$, after applying $\mbox{Hom}(-,A)$: $0\leftarrow T(M)\leftarrow P_1^*\overset{\mbox{Hom}(\partial,A)}\longleftarrow P_0^*\leftarrow M^*\leftarrow0$ How does one show that $T(M)$ is independent, up to projective equivalence, on the choice of the projective resolution? Many books mention this fact but none of them seem to give a proof.
Proving projective equivalence of Auslander Transpose
3
$\begingroup$
abstract-algebra
commutative-algebra
homological-algebra
-
1@Theo Buehler Yes, I saw that hint, but as is often the case with Eisenbud, I don't see how that can be of any help. – 2011-09-16
1 Answers
1
I found an answer here: