Property of transpose Tr in category of modules (Auslander-Reiten theory)

I am reading the book Elements of the Representation Theory of Associative Algebras: Volume 1 , and I have a question on its proof of main properties of transpose $$Tr$$, appearing in Auslander-Reiten theory.
Here we assume $$A$$ is a finite dimensional algebra over an algebraically closed field $$K$$, and $$mod A$$ is a category of finitely generated modules of $$A$$. Also, consider a functor $$(−)^t = Hom_A(−,A) : mod A → mod A^{op}$$.
For an object $$M$$ in $$mod A$$, we define $$Tr M$$ as follows:

If $$P_1→P_0→M→0$$ is a minimal projective presentation, then $$Tr M=Coker(P_0^t→P_1^t)$$. In other words, $$P_0^t→P_1^t→TrM→0$$ is exact.

Proposition 2.1.(b) in the textbook p.107 is about showing that

If M is not projective, then the exact sequence $$P_0^t→P_1^t→TrM→0$$ is a minimal projective presentation of an $$A^{op}$$ module $$Tr M$$.

In the proof, author assumes $$P_0^t→P_1^t→TrM→0$$ is not a minimal projective presentation. Then there exist non-trivial decompositions (by projective modules) $$P_0^t=E_0'⊕E_0''$$, $$P_1^t=E_1'⊕E_1''$$ and an isomorphism $$v:E_0''→E_1''$$ and a homomorphism $$u:E'_0→E'_1$$ such that the previous exact sequence is expressed as $$E_0'⊕E_0''\overset{\begin{pmatrix} u & 0 \\ 0 & v \end{pmatrix}}{\rightarrow}E_1'⊕E_1''→Tr M→0$$.
(I think the author assumed $$E'_0\overset{u}{\rightarrow}E'_1→Tr M→0$$ is a minimal projective presentation.)

But I couldn't quite understand the next step. It is written as,

But then applying $$(−)^t$$ yields a projective presentation of M of the form $${E'_1}^t\overset{u^t}{\rightarrow}{E'_0}^t→M→0$$

But I couldn't figure out why we get $$M$$ as a cokernel of $$u^t$$. Any helps will be appreciated. Thank you for reading.

For finitely generated projective modules $$X$$, the evaluation map $$X\to X^{tt}$$ (defined by $$x\mapsto[\varphi\mapsto\varphi(m)]$$) is a natural isomorphism.
So $$P_1\to P_0$$ is isomorphic to $$P_1^{tt}\to P_0^{tt}$$, which in turn is isomorphic to the direct sum of $$E_1'^t\xrightarrow{u^t}E_0'^t$$ and $$E_1''^t\xrightarrow{v^t}E_0''^t$$.
But $$v$$ is an isomorphism, so $$v^t$$ is an isomorphism and so its cokernel is zero. So $$M$$, which is the cokernel of $$P_1\to P_0$$, is also the cokernel of the first summand $$E_1'^t\xrightarrow{u^t}E_0'^t$$.