A question on Auslander-Bridger transpose I am learning Auslander-Reiten Theory. When I read the book Frobenius Algebras I. Basic Representation Theory, I have some problems.
On page 236-237, there is the following

Proposition 4.5. Let $M$ and $N$ be modules in $\operatorname{mod} A$. The following statements hold.
  (i) $\operatorname{Tr}(M)=0$ if and only if $M$ is from $\operatorname{proj}A$.
  (ii) $\operatorname{Tr}(M)$ is a module in $\operatorname{mod}_{\mathscr P} A^{op}$.
  (iii) If $M$ is from $\operatorname{mod}_{\mathscr P} A$ and
  $$P_1 \overset{p_1}\to P_0 \overset{p_0}\to M \to 0$$
  is a minimal projective presentation of $M$ in $\operatorname{mod}A$, then the induced exact sequence
  $$P_0^t \overset{p_1^t}\to P_1^t \overset{\pi_M}\to \operatorname{Tr}(M)\to 0$$
  is a minimal presentation of $\operatorname{Tr}(M)$ in $\operatorname{mod} A^{op}$.

Proof.

I can't figure out how to get the commutative diagram. Can anyone help me?
 A: @Amit I still don't think it's correct. If you actually name the morphisms and try to prove it, you will see it just doesn't gives. I will start right where we have half of the problem completed (that is, the first part of desired diagram complete).
$$ \require{AMScd}\begin{CD} 
P_{1}^{t} @>\pi_{M}>> \text{Tr}M @>>> 0\\
@VV\psi V \circ @|\\
U_{1}\oplus V_{1} @>(w\;0)>> \text{Tr}M @>>> 0
\end{CD}$$
with $\psi$ an isomorphism.
Let $\psi_1$ be the projection of $\psi$ onto the first coordinate $U_1$ and consider the composition $\psi_1 \circ p_1^t: P_0^t \rightarrow Ker(w)$. It is well defined since $w \circ \psi_1 \circ p_1^t = (w,0) \circ \psi \circ p_1^t = \pi \circ p_1^t = 0$ and is easy to see that is surjective. So by definition of projective cover there exists a surjective morphism $\psi': P_0^t \rightarrow U_0$ such that $ u \circ \psi' = \psi_1 \circ p_1^t$. I think so far we are in the same route. As you say, since $\psi'$ is surjective, there exists a retraction $ r: U_0 \rightarrow P_0^t$ such that $ \psi' \circ r = Id_{U_0}$.
That implies that $P_0^t =  Img(r) \oplus Ker(\psi')$ (that's an equality of sets). So in the end our setup is
$$ \require{AMScd}\begin{CD} 
Img(r) \oplus Ker(\psi') @>p_1^t>> P_1^t\\
@VV\psi ' \oplus Id V  @VV\psi V\\
U_0\oplus Ker(\psi') @> u \ \oplus \  \left.\psi_2 \circ p_1^t\right|_{Ker(\psi')}>> U_{1}\oplus V_{1}
\end{CD}$$
where the morphism $\left.\psi_2 \circ p_1^t\right|_{Ker(\psi')}$ is the only one from $Ker(\psi')$ to $V_1$ that makes sense. 
But the diagram doesn't seem to conmute:  in the upper right side you get $\psi \circ p_1^t (x \oplus y) = \psi_1 p_1^t (x\oplus y) +  \psi_2 p_1^t (x\oplus y) $ with $x \in Img(r)$ and $y \in ker(\psi')$ . Since $y \in Ker(\psi ')$, $ \psi \circ p_1^t (x \oplus y)  = \psi_1 p_1^t (x) +  \psi_2 p_1^t (x\oplus y)$.
But the composition on the down left side you get  $  \psi_1 p_1^t (x) +  \psi_2 p_1^t (y)$
A: We can always find a minimal projective presentation $U_{0}\overset{u}{\to}U_{1}\overset{w}{\to}\text{Tr}(M)\to 0$ of $\text{Tr}(M)$ (take a projective cover of $\text{Tr}(M)$, then a projective cover of the kernel of that epimorphism). So then we have the following setup: $$ \require{AMScd}
\begin{CD}
     @. P_{1}^{t}\\
    @. @VV \pi_{M} V\\
    U_{1} @>>w> \text{Tr}(M) @>>> 0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{exact}\\ \end{CD}$$ with an exact row as indicated. As $P_{1}^{t}$ is projective we get the existence of a lift of $\pi _{M}$ to $U_{1}$. That is, there exists a map $\psi \colon P_{1}^{t}\to U_{1}$ such that $\pi_{M}=w\circ \psi$. As $w$ is the morphism in a projective cover of $\text{Tr}(M)$ and since $w\circ \psi=\pi_{M}$ is surjective, we must have that $\psi$ is also surjective (see, for example, p.28 in Elements of Representation Theory of Associative Algebras Vol. 1 by Assem, Simson and Skowronski). Therefore, we get a short exact sequence $$ \require{AMScd}
\begin{CD}
0@>>>\text{ker}(\psi) @>i>> P_{1}^{t} @>\psi>> U_{1} @>>> 0
\end{CD}$$
which must split as $U_{1}$ is projective. Thus, $P_{1}^{t}\cong U_{1}\oplus V_{1}$, where $V_{1}\cong \text{ker}(\psi)$.
This yields the following commutative diagram:
$$ \require{AMScd}\begin{CD} 
P_{1}^{t} @>\pi_{M}>> \text{Tr}M @>>> 0\\
@VVV \circ @|\\
U_{1}\oplus V_{1} @>(w\;0)>> \text{Tr}M @>>> 0
\end{CD}$$
where the map on the left isn't really the identity on $P_{1}^{t}$ but the isomorphism induced by the splitting detailed above. (The authors of your book have been a bit sloppy there.)
Since $U_{0} \to U_{1} \to \text{Tr}M \to 0$ is a minimal projective presentation, we have that there is a surjection $U_{0} \to \ker(w)$. In particular, since the map $P_{1}^{t} \to U_{1}\oplus V_{1}$ is an isomorphism, we know $\ker \pi_{M} = \ker w \oplus V_{1}$. The map $p_{1}^{t}$ factors through (and surjects onto it as $\text{im}p_1^{t} = \ker \pi_M$) $\ker\pi_{M}$ (since the exact sequence $$ P_{0}^{t} \to P_{1}^{t} \to \text{Tr}M\to 0$$ is induced by a minimal projective presentation of $M$), so we have the following: $$\begin{CD}
     @. P_{0}^{t}\\ 
@. @VV g V \\
     @. \ker \pi_{M}=\ker w \oplus V_{1}\\
@. @VV f V\\
U_{0} @>>w'> \ker w@>>> 0,\end{CD}$$
where $f,g$ are both surjective. Since $P_0^t$ is projective, we get a lift of $fg$ through $w'$, and (as before) since $w'$ is a projective cover we must have that the lift $\psi ' \colon P_0^t \to U_0$ is surjective. The induced short exact sequence splits and $P_0^t \cong U_0 \oplus V_0$ where $V_0 = \ker \psi '$. Finally, recall that $U_0\oplus V_0 \cong P_0^t$ surjects onto $\ker \pi_{M} = \ker w \oplus V_{1}$ and $U_0$ only maps into $\ker w$, so we must have that $V_0$ surjects onto $V_1$.
