- #1
Math Amateur
Gold Member
MHB
- 3,920
- 48
I am reading Adhikari and Adhikari's (A&A) book, "Basic Modern Algebra with Applications".
I am currently focussed on Section 9.7 Exact Sequences.
On page 391 A&A state and prove the Short Five Lemma. I need help with some of the details of the 'diagram chasing' in the proof.
The Short Five Lemma and the first part of its proof read as follows:https://www.physicsforums.com/attachments/3628
https://www.physicsforums.com/attachments/3629In the first two lines of the above proof we find the following:" ... Suppose $$\beta (b) = 0$$ for some $$b \in B$$. We shall show that $$b = 0$$.
Now $$\gamma g (b) = g' \beta (b) = 0 \Longrightarrow g(b) = 0$$ since $$\gamma$$ is a monomorphism ... ... "
Now I need someone to critique my detailed reasoning reasoning concerning these statements - I think I understand ... but then, I am working by myself on this material ... so a confirmation that I am on the right track would be most helpful ...Now ... my reasoning is as follows:
We suppose that $$\beta (b) = 0_{B'} $$
We need to show that $$b = 0_B$$ ... ...
... which implies that $$\text{ker } \beta = 0_B$$ ... ...
... which implies that $$\beta$$ is an injective homomorphism ... ... that is a monomorphism ...Now $$ \gamma g (b) = g' \beta (b) $$ by the commutativity of the diagram (Fig. 9.7)But $$g' \beta (b) = g' ( 0_{B'} ) = 0_{C'}$$ since $$g'$$ is a homomorphism ...So we have $$\gamma g (b) = \gamma ( g (b) ) = 0_{C'}$$But then $$\gamma$$ is a monomorphism, so that the only element $$x$$ in its domain $$C$$ that gives $$\gamma (x) = 0_{C'}$$ is $$x = 0_{C'}$$ ...So then we have $$g(b) = 0_{C'}$$ ...
... ... and then the proof of (i) continues ...
Can someone please confirm that the details of my analysis above regarding the first statements of the proof is correct and/or critique my analysis pointing out any errors or shortcomings ... ...
Peter
I am currently focussed on Section 9.7 Exact Sequences.
On page 391 A&A state and prove the Short Five Lemma. I need help with some of the details of the 'diagram chasing' in the proof.
The Short Five Lemma and the first part of its proof read as follows:https://www.physicsforums.com/attachments/3628
https://www.physicsforums.com/attachments/3629In the first two lines of the above proof we find the following:" ... Suppose $$\beta (b) = 0$$ for some $$b \in B$$. We shall show that $$b = 0$$.
Now $$\gamma g (b) = g' \beta (b) = 0 \Longrightarrow g(b) = 0$$ since $$\gamma$$ is a monomorphism ... ... "
Now I need someone to critique my detailed reasoning reasoning concerning these statements - I think I understand ... but then, I am working by myself on this material ... so a confirmation that I am on the right track would be most helpful ...Now ... my reasoning is as follows:
We suppose that $$\beta (b) = 0_{B'} $$
We need to show that $$b = 0_B$$ ... ...
... which implies that $$\text{ker } \beta = 0_B$$ ... ...
... which implies that $$\beta$$ is an injective homomorphism ... ... that is a monomorphism ...Now $$ \gamma g (b) = g' \beta (b) $$ by the commutativity of the diagram (Fig. 9.7)But $$g' \beta (b) = g' ( 0_{B'} ) = 0_{C'}$$ since $$g'$$ is a homomorphism ...So we have $$\gamma g (b) = \gamma ( g (b) ) = 0_{C'}$$But then $$\gamma$$ is a monomorphism, so that the only element $$x$$ in its domain $$C$$ that gives $$\gamma (x) = 0_{C'}$$ is $$x = 0_{C'}$$ ...So then we have $$g(b) = 0_{C'}$$ ...
... ... and then the proof of (i) continues ...
Can someone please confirm that the details of my analysis above regarding the first statements of the proof is correct and/or critique my analysis pointing out any errors or shortcomings ... ...
Peter
Last edited: