A lemma from Hilton & Stammbach's book A Course in Homological Algebra In orde to prove the set of equivalence classes of extensions of $A$ by $B$ is a contravariant functor of the first component and covariant functor of the second. The authors give us three lemma. I'm confused with lemma 1.3. Here is the several things I don't understand.


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*Why we obtain an extension $B \stackrel{\alpha}{\rightarrowtail} P \stackrel{\beta} {\twoheadrightarrow} A'$ in the proof of lemma 1.3?

*What means $\zeta$ induces identity both in $A'$ and $B$. And how can I say $\zeta$ is an isomorphism by lemma I.1.1?


PS: I have to post the picture since the $\LaTeX$ command \xymatrix doesn't works here.
Here is the lemma I.1.1 from the Chapter I.

And this is the lemma III.1.2 and lemma III.1.3. 

 A: Since $\epsilon$ is an epimorphism, there is an extension $\ker(\epsilon)\rightarrowtail P\stackrel{\epsilon}\twoheadrightarrow A'$, but there is an isomorphism $B\to \ker(\epsilon)$, so composing with this isomorphism we get an extension $B\rightarrowtail P\stackrel{\epsilon}\twoheadrightarrow A'$.
For your second question, I agree that this wording is rather unclear and in fact some steps are left out.  At this point in the argument, you have two different short exact sequences that start with $B$ and end in $A'$, namely $B\stackrel{\kappa'}\rightarrowtail E'\stackrel{\nu'}\twoheadrightarrow A'$ and $B\stackrel{\mu}\rightarrowtail P\stackrel{\epsilon}\twoheadrightarrow A'$.  Furthermore, I claim that $\mu=\zeta\kappa'$.  Indeed, $\mu$ is defined as the unique map $B\to P$ such that $\varphi\mu=\kappa$ and $\epsilon\mu=0$, and $\zeta\kappa'$ satisfies this definition since $\varphi\zeta\kappa'=\xi\kappa'=\kappa$ and $\epsilon\zeta\kappa'=\nu'\kappa'=0$.  We also know that $\epsilon\zeta=\nu'$.  This means that $\zeta$ and our two short exact sequences fit into a commutative diagram
$$\require{AMScd}
\begin{CD}
B @>{\kappa'}>> E' @>{\nu'}>> A'\\
@V{1}VV @V{\zeta}VV @V{1}VV\\
B @>{\mu}>> P @>{\epsilon}>> A'
\end{CD}$$
where the other two vertical maps are the identity maps.  Since the identity maps are isomorphisms, Lemma I.1.1 gives that $\zeta$ is an isomorphism.
