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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!

Quiet_waters
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1 Answers1

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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.

N.B.
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  • Sorry, I am with troubles and difficulties: where did I use $\overline{v}$? I just could see that I always will have $d(v(x))=0$ and always $d(x'')$ is at form $d(v(x))$, so I just could see $d\equiv 0$. I know I am wrong and you right, but I could not understand. Could you help me once again please? Many thanks. – Quiet_waters Oct 01 '19 at 13:50
  • Once the lines are exacts, $v$ is surjective, so all element of $ker(f'')$ is at form $v(x)$, what do you think? Thank you very much. – Quiet_waters Oct 01 '19 at 14:12
  • Yes... $\overline{u}'$ is not injective. Many thanks! – Quiet_waters Oct 01 '19 at 14:33