A little problem on exact sequences, part 2.

[steenis]

Part 1 of this little problem is here: http://mathhelpboards.com/linear-abstract-algebra-14/little-problem-exact-sequences-19368.html.

This is part 2: who can formulate and prove the dual statement?

 

[Deveno]

You should clarify what you mean by "dual", which might mean:

a) The statement obtained by "reversing all the arrows", or

b) the statement obtained by considering the "dual modules" $\text{Hom}_R(-,R)$.

 

[steenis]

I am afraid it is neither a) nor b), so I doubt if I used the term "dual" correctly. But it is about the dual notions ker and coker. This is what I know of the solution:We have two R-maps $f:A\to B$ and $g:B\to C$ of left R-modules.
And we have an isomorphism $A\cong \ker g$ "induced by f" .
Then prove that the sequence $S:0\to A\to _f B\to _g C$ is left-exact, i.e.,
$\mbox{im} f = \ker g$ and $f$ is injective.Proof:
Let $\overline f : A\to \ker g : a\mapsto f(a)$ be the isomorphism between $A$ and $\ker g$ induced by f. This implies that $\mbox{im }f \subset \ker g$ and the well-definition of $\overline f$ (because $f$ is well-defined).
Subsequently, we have:
$x\in \ker g \Leftrightarrow \exists (a\in A) \mbox{ } \overline f (a) = f(a) = x \Leftrightarrow x\in \mbox{im }f$, i.e., $\mbox{im }f = \ker g$.
And:
$x\in \ker f \Rightarrow x\in \ker \overline f \Rightarrow x=0$, i.e., $f$ is injective. $\Box$

Your comments, please.

 

[Deveno]

I am afraid it is neither a) nor b), so I doubt if I used the term "dual" correctly. But it is about the dual notions ker and coker. This is what I know of the solution:We have two R-maps $f:A\to B$ and $g:B\to C$ of left R-modules.
And we have an isomorphism $A\cong \ker g$ "induced by f" .
Then prove that the sequence $S:0\to A\to _f B\to _g C$ is left-exact, i.e.,
$\mbox{im} f = \ker g$ and $f$ is injective.Proof:
Let $\overline f : A\to \ker g : a\mapsto f(a)$ be the isomorphism between $A$ and $\ker g$ induced by f. This implies that $\mbox{im }f \subset \ker g$ and the well-definition of $\overline f$ (because $f$ is well-defined).
Subsequently, we have:
$x\in \ker g \Leftrightarrow \exists (a\in A) \mbox{ } \overline f (a) = f(a) = x \Leftrightarrow x\in \mbox{im }f$, i.e., $\mbox{im }f = \ker g$.
And:
$x\in \ker f \Rightarrow x\in \ker \overline f \Rightarrow x=0$, i.e., $f$ is injective. $\Box$

Your comments, please.

That is "dual" as in a) of my previous post. Note that in part 1), $B/(\text{im }f)$ *is* $\text{coker }f$, so that the dual of the diagram:

$A \stackrel{f}{\to}B \stackrel{g}{\to}C \cong \text{coker }f \to 0$ is:

$0 \to C\cong \text{ker }k\stackrel{h}{\to}B\stackrel{k}{\to}A$

(I have labelled the reverse arrows $C \to B$ and $B \to A$ by $h$ and $k$ to avoid thinking of them as "inverse functions"), with the isomorphisms *induced* by the arrow just before on part 1), and just after in part 2).

 

[steenis]

Thank you, Deveno, very clear.