I Direct Products of Modules ....Another Question .... ....

[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:

In 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:

Hope that helps ...

Peter

 

[fresh_42]

When Bland puts $g(x) = ( x_\alpha )$ ...

he is writing down some (any) homomorphism $g$. It has $x$ as input and an element of the direct product as output. Those elements are of the form $(x_\alpha )\,.$ He does not specify

... a particular $g$.

Particular only in so far, as he has only one $g$, but this always looks as described. He did not make any assumptions on the $x_\alpha$ other than $x_\alpha \in M_\alpha$.

It's like saying: Consider a function $g: \mathbb{R} \longrightarrow \mathbb{R}^2$ given by $g(x)=(x_1,x_2)$. Does it tell you how the function works? It only says $g: \mathbb{R} \longrightarrow \mathbb{R}^2\,.$

 

[Math Amateur]

he is writing down some (any) homomorphism $g$. It has $x$ as input and an element of the direct product as output. Those elements are of the form $(x_\alpha )\,.$ He does not specify
Particular only in so far, as he has only one $g$, but this always looks as described. He did not make any assumptions on the $x_\alpha$ other than $x_\alpha \in M_\alpha$.

It's like saying: Consider a function $g: \mathbb{R} \longrightarrow \mathbb{R}^2$ given by $g(x)=(x_1,x_2)$. Does it tell you how the function works? It only says $g: \mathbb{R} \longrightarrow \mathbb{R}^2\,.$

Thanks fresh_42 ... that clarifies that issue ...

Appreciate your help ...

Peter