Split Short Exact Sequences - Bland - Proposition 3.2.6

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am trying to understand Section 3.2 on exact sequences in Mod_R and need help with the proof of Proposition 3.2.6.

Proposition 3.2.6 and its proof read as follows:View attachment 3609The part of the proof that perplexes me is the section of the proof that begins with the claim that $$g'$$ is well defined. That section of the proof reads as follows:https://www.physicsforums.com/attachments/3610Now Bland takes two elements from $$Ker f'$$, namely:

$$(x - f(f'(x)))$$ and $$(x - f(f'(x))) $$

and forms their difference (which must also belong to $$Ker f'$$ since $$Ker f'$$ is a submodule) thus forming:

$$(x - f(f'(x))) - (x - f(f'(x)))$$

which Bland shows is equal to

$$(x - x') - f(f'(x - x'))
$$Now he has already shown that $$(x - x') \in I am f$$ and certainly $$f(f'(x - x')) \in I am f$$ ... ...

So we have that:

$$(x - x') - f(f'(x - x')) \in Ker f' \cap I am f $$

and also we have

$$Ker f' \cap I am f = 0$$ since $$M = Ker f' \oplus I am f$$

... ...

BUT ... ...

... ... why does this prove that $$g'$$ is "well defined" ... indeed what does Bland mean by this?

Further to the question can someone please confirm that my analysis above is correct ... or better ... critique my analysis pointing out any questionable or inaccurate statements ...

Help will be appreciated ...

Peter

 

[Fallen Angel]

Hi Peter,

Bland's has just constructed an application $g'$, that may not be well defined because he tooks a definition based in $g'(y)=...(x)$ with $g(x)=y$.

This definition may depend on $x$, because $g$ doesn't need to be injective.

So he needs to prove that if you choose $x,x'$ with $g(x)=g(x')=y$ then $g'(y)=...(x)=...(x')$. That is, it doesn't mind what g-pre-image you choose for the element $y$, $g'(y)$ will be the same.

Now what he tooks is two elements such that $g(x)=g(x')$, and compute
$...(x)-...(x')=(x-f(f'(x)))-(x'-f(f'(x')))$ and proves that this difference lay on $Ker f' \cap I am f=0$, so $(x-f(f'(x)))-(x'-f(f'(x')))=0$, then $...(x)=...(x')$ and $g'$ is well defined.

 

[Math Amateur]

Hi Peter,

Bland's has just constructed an application $g'$, that may not be well defined because he tooks a definition based in $g'(y)=...(x)$ with $g(x)=y$.

This definition may depend on $x$, because $g$ doesn't need to be injective.

So he needs to prove that if you choose $x,x'$ with $g(x)=g(x')=y$ then $g'(y)=...(x)=...(x')$. That is, it doesn't mind what g-pre-image you choose for the element $y$, $g'(y)$ will be the same.

Now what he tooks is two elements such that $g(x)=g(x')$, and compute
$...(x)-...(x')=(x-f(f'(x)))-(x'-f(f'(x')))$ and proves that this difference lay on $Ker f' \cap I am f=0$, so $(x-f(f'(x)))-(x'-f(f'(x')))=0$, then $...(x)=...(x')$ and $g'$ is well defined.

Thanks for you help, Fallen Angel ... ...

Still reflecting on what you have written ...

Will get back to you soon on this matter ...

Thanks again,

Peter