Boundary Homomorphism I was studying the proposition 2.10 of Atiyah and MacDonald's Introduction to Commutative Algebra, and have a question.
The proposition says:

Let
  $$
\require{AMScd}
\begin{CD}
0 @>>> M^{'} @>{u}>> M @>{v}>> M^{''} @>>> 0 \\
@. @V{f^{'}}VV @V{f}VV @VV{f^{''}}V \\
0 @>>> N^{'} @>{u^{'}}>> N @>{v^{'}}>> N^{''} @>>> 0
\end{CD}$$
  be a comutative diagram of $A$-modules and homomorphisms, with the rows exact. Then exists an exact sequence:
  $$\cdots\stackrel{\bar{v}}\longrightarrow Ker(f^{''})\stackrel{d}\longrightarrow Coker(f^{'})\stackrel{\bar{u}^{'}}\longrightarrow\cdots.$$
  Where the boundary homomorphism $d$ is defined as follows: if $x^{''} \in Ker(f^{''})$, we have $x^{''}=v(x)$ for some $x \in M$, and $v^{'}(f(x))=f^{''}(v(x))=0$, hence $f(x) \in Ker(v^{'})=Im(u^{'})$, so that $f(x)=u^{'}(y^{'})$ for some one $y^{'} \in N^{'}$. Then $d(x^{''})$ is defined to be the image of $y^{'}$ in $Coker(f^{'})$.

My ask is if $v^{-1}(\{x^{''}\})$ have more than one element, then $d$ it's well defined?
I searched about the boundary homomorphism and I found another definition like this:
Interpretation of boundary homomorphism in long exact sequence of homology groups .
Have any connection?
Thanks.
 A: In order to show that $d$ is well-defined, we need to show that $d(x'')$ doesn't depend on what element of $v^{-1}(x'')$ we choose. 
We've already chosen $x\in v^{-1}(x'')$, so suppose that $v(z)=x''$ for some other $z\in M$. Then by the same reasoning as in the book, $f(z)=u'(w')$ for some $w'\in N'$.
For $d$ to be well-defined, we need
$$d(x'')=\text{ image of }y'\text{ in }\mathrm{coker}(f')=\text{ image of }w'\text{ in }\mathrm{coker}(f')$$
The image of $y'$ and $w'$ in $\mathrm{coker}(f')$ are equal iff $y'-w'\in\mathrm{im}(f')$, by the definition of cokernel.
Now consider:
$$\begin{align*}
y'-w'\in\mathrm{im}(f')&\iff u'(y')-u'(w')\in\mathrm{im}(u'\circ f')\text{ because }u'\text{ injective}\\\\
&\iff u'(y')-u'(w')\in\mathrm{im}(f\circ u) \text{ because }u'\circ f'=f\circ u\\\\
&\iff f(x)-f(z)\in\mathrm{im}(f\circ u)\text{ by definition of }y',w'\\\\
&\iff x-z\in\mathrm{im}(u)\\\\
&\iff x-z\in\ker(v)\text{ because }\mathrm{im}(u)=\ker(v)\\\\
&\iff v(x)=v(z)
\end{align*}$$
and $v(x)=v(z)=x''$ because that's how we chose $x$ and $z$ in the first place. Thus no matter what element of $v^{-1}(x'')$ we choose, the definition of $d$ gives the same answer.
