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

In summary, Peter is reading Paul E. Bland's book "Rings and Their Modules" and is currently focused on Section 3.2 Exact Sequences in \text{Mod}_R. He needs help understanding Definition 3.2.2 and Proposition 3.2.3, which state that $f'f = \text{id}_{M_{1}}$ and $f'(x) \in M_{1}$ for all $x \in M$. Even when $x \in M - f(M_{1})$, $f'(x)$ still exists because $f'$ is defined on all of $M$. Peter is grateful for the explanation of how Definition 3.2.2 works.
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I am reading Paul E. Bland's book "Rings and Their Modules" ...

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

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

Peter
 
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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 $M$.
 
  • #3
GJA said:
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
 

1. What is Split Monomorphisma?

Split Monomorphisma is a mathematical concept that refers to a type of group homomorphism in abstract algebra. It describes a function that maps elements of one group to elements of another group in a way that preserves the group structure.

2. What does "Bland Definition 3.2.2" mean in relation to Split Monomorphisma?

"Bland Definition 3.2.2" refers to a specific definition of Split Monomorphisma that was proposed by mathematician Richard Bland. This definition helps to clarify the concept and provide a clear understanding of its properties and characteristics.

3. What is Proposition 3.2.3 in relation to Split Monomorphisma?

Proposition 3.2.3 is a mathematical statement that was proven by mathematician Richard Bland about the properties of Split Monomorphisma. It provides a useful theorem that can be used to prove other theorems and properties related to this concept.

4. How is Split Monomorphisma used in scientific research?

Split Monomorphisma is a fundamental concept in abstract algebra and is used in a variety of fields within mathematics and science. It is particularly useful in group theory, which has applications in fields such as physics, chemistry, and computer science.

5. Are there any real-world applications of Split Monomorphisma?

Yes, there are many real-world applications of Split Monomorphisma. For example, it is used in cryptography to create secure communication systems, in chemistry to understand the structure of molecules, and in computer science to design efficient algorithms for data processing and storage.

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