I am dealing with this proposition (Boundary Homomorphism):
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 $Coker(f^{'})$.
And have the following doubty:
$d(v(x))=0\iff\\ y'\in Im(f')\iff \\\overline{u}'(y'+ Im(f'))=0\iff \\u'(y')+Im(f)=0\iff \\u'(y')=f(x)$
for some $x \in M.$
But this holds for any $v(x),$ so could $d$ be the null homomorphism...?
Many thanks in advance!
Disregard my previous answer: the problem is in the second logic equality of your reasoning. For one of the implications you are using that $\overline{u}'$ must be injective which is false, even if the starting $u'$ is mono. If you consider the derived long exact sequence injectivity of $\overline{u}'$ is equivalent to $d$ being the zero map as you deduce.