MHB Split Monomorphisma .... Bland Definition 3.2.2 & Proposition 3.2.3 ....

[Math Amateur]

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

Currently I am focused on Section 3.2 Exact Sequences in $$\text{Mod}_R$$ ... ...

I need some help in order to fully understand Definition 3.2.2 and Proposition 3.2.3 ...

Definition 3.2.2 and Proposition 3.2.3 read as follows:
https://www.physicsforums.com/attachments/8072
In Definition 3.2.2 we read that $$f'f = \text{id}_{M_1}$$ ... ... BUT ... ... I thought that $$f'f$$ was only defined on $$f(M) = \text{Im } f $$ ... ... what then happens to elements $$x \in M$$ that are outside of $$f(M) = \text{Im } f$$ ... ... see Fig. 1 below ...
View attachment 8073Note that in the proof of Proposition 3.2.3 we read:" ... ... If $$x \in M$$, then $$f'(x) \in M_1$$ ... ... " But ... how does this work for $$x$$ outside of $$f(M) = \text{Im } f $$ such as $$x$$ shown in Fig. 1 above?
I would be grateful if someone could explain how Definition 3.2.2 "works" ... ...

Peter

 

[GJA]

Hi, Peter.

The condition that needs to occur in order to form the composition $f'f$ (or any function composition) is that $f$'s range must be a subset of $f'$'s domain. Since $f'$ is defined on all of $M$, there is no issue forming $f'f$.

Even when $x\in M-f(M_{1})$, $f'(x)$ still exists because $f'$ is defined on all of $M$.

 

[Math Amateur]

Hi, Peter.

The condition that needs to occur in order to form the composition $f'f$ (or any function composition) is that $f$'s range must be a subset of $f'$'s domain. Since $f'$ is defined on all of $M$, there is no issue forming $f'f$.

Even when $x\in M-f(M_{1})$, $f'(x)$ still exists because $f'$ is defined on all of $M$.

Thanks GJA ...

Appreciate your help...

Peter