MHB Does Proposition 3.2.6 Imply g'(y) = x' - f(f'(x'))?

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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 $$\text{Mod}_R$$ ... ...

I need some further help in order to fully understand the proof of Proposition 3.2.6 ...

Proposition 3.2.6 and its proof read as follows:
View attachment 8079
In the above proof of Proposition 3.2.6 we read the following:"... ... now define $$g' \ : \ M_2 \longrightarrow M$$ by $$g'(y) = x - f(f'(x))$$, where $$x \in M$$ is such that $$g(x) = y$$ ... ... ... ...

... ... Suppose that $$x' \in M$$ is also such that $$g(x') = y$$ ... ... Does the above text imply that $$g'(y) = x' - f( f'(x') )$$ ... ... ?

Peter
 
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Hi Peter,

Yes, indeed, $g'(y) = x-f(f'(x)) = x' - f(f'(x'))$, since the last equality has just been proved. This shows that $g'(y)$ in unambiguously defined; please refer to my previous post for an intuitive explanation.
 

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