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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.

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Any two projective resolutions are chain homotopic. Does this help? –  Rasmus Sep 16 '11 at 14:43
    
@Rasmus $\mbox{Ext}^1(M,A)\subset T(M)$, but I don't see how isomorphism of $\mbox{Ext}^1(M,A)$ leads to projective equivalence of $T(M)$? –  ashpool Sep 16 '11 at 15:06
    
I haven't really thought about this, but here's a hint from Eisenbud's commutative algebra: reduce to the case where the comparison maps in two projective resolutions of $M$ are surjective. –  t.b. Sep 16 '11 at 15:10
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@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. –  ashpool Sep 16 '11 at 15:16

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