MHB Understanding Proposition 2.1.1 in Paul E. Bland's Rings & Modules

[Math Amateur]

I am reading Paul E. Bland's book: Rings and Their Modules and am currently focused on Section 2.1 Direct Products and Direct Sums ... ...

I need help with another aspect of the proof of Proposition 2.1.1 ...

Proposition 2.1.1 and its proof read as follows:
https://www.physicsforums.com/attachments/8032In the above proof by Paul Bland we read the following:

" ... ... suppose that $$g \ : \ N \rightarrow \prod_\Delta M_\alpha$$ is also an $$R$$-linear mapping such that $$\pi_\alpha g = f_\alpha$$ for each $$\alpha \in \Delta$$. If g(x) = ( x_\alpha ) ... ... "When Bland puts $$g(x) = ( x_\alpha )$$ he seems to be specifying a particular $$g$$ and then proves $$f = g$$ ... ... I thought he was proving that for any g such that $$\pi_\alpha g = f_\alpha$$ we have $$f = g$$ ... can someone please clarify ... ?Help will be much appreciated ... ...Peter
======================================================================================The above post mentions but does not define $$f$$ ... Bland's definition of $$f$$ is as follows:
View attachment 8033
Hope that helps ...

Peter

 

[steenis]

Bland is saying that any $g$ that satisfies the condition $\pi_\alpha \circ g = f_\alpha$ must be equal to $f$.
So take a $g$ that satisfies the condition and prove that $g=f$, that’s all.

So take such a $g$ then $g$ is an R-map $g:N \longrightarrow \prod_\Delta M_\alpha$, take $x\in N$, then $g(x) \in \prod_\Delta M_\alpha$.
This means that $g(x)$ has the form $g(x)=(x_\alpha)$, where $x_\alpha$ is the coordinate of $g(x)$ in $M_\alpha$, for all $\alpha \in \Delta$, and so on.