Start on Bland Problem 1, Problem Set 4.1: Generating & Cogenerating Modules

[steenis]

Very good, Peter, I knew you can do it !A very small comment, the last lines are a little bit confusing and not quite correct.
I would write that something like this

Let $y \in N = \sum_\Delta \text{ I am } h_\alpha$.

Then there are $x_\alpha \in M$ for all $\alpha \in \Delta$, such that $y = \sum_\Delta h_\alpha(x_\alpha)$.

Let $x = (x_\alpha)_\Delta$, then $x \in M^{(\Delta)}$ and $f(x) =f((x_\alpha)) = \sum_\Delta h_\alpha(x_\alpha)$ = y.

Therefore $f$ is an epimorpism.

 

[Math Amateur]

Very good, Peter, I knew you can do it !A very small comment, the last lines are a little bit confusing and not quite correct.
I would write that something like this

Let $y \in N = \sum_\Delta \text{ I am } h_\alpha$.

Then there are $x_\alpha \in M$ for all $\alpha \in \Delta$, such that $y = \sum_\Delta h_\alpha(x_\alpha)$.

Let $x = (x_\alpha)_\Delta$, then $x \in M^{(\Delta)}$ and $f(x) =f((x_\alpha)) = \sum_\Delta h_\alpha(x_\alpha)$ = y.

Therefore $f$ is an epimorpism.

Thanks for all the help, Steenis ... thanks to you I understand a lot more than when I started working on the problem ...

Regarding the last few lines of the above proof I thought I'd write out explicitly what is going for the case \(\displaystyle \Delta = \{ 1,2, \ ... \ ... \ , \ n \}\)

We have \(\displaystyle f \ : \ M^{ ( \Delta ) } = \bigoplus_\Delta M \rightarrow N\) where \(\displaystyle f = \sum_H g = \sum_\Delta h_\alpha\) ... and \(\displaystyle h_\alpha \ : \ M \rightarrow N\) ... ...

... and so ... \(\displaystyle f(( x_\alpha )) = \sum_\Delta h_\alpha ( x_\alpha )\) ...
Now to show or illustrate explicitly in the case of \Delta = \{ 1,2, \ ... \ ... \ , \ n \} how f is an epimorphism ... ( ... not meant to be a proof ... just an illustration ...)Assume \(\displaystyle y_1 \in \text{Im } h_1 \Longrightarrow \exists \ x_1\) such that \(\displaystyle h_1 ( x_1 ) = y_1\) ...

and ... \(\displaystyle y_2 \in \text{Im } h_2 \Longrightarrow \exists \ x_2\) such that \(\displaystyle h_2 ( x_2 ) = y_2\)

... ... ... ... ...

... ... ... ... ...

and ... \(\displaystyle y_n \in \text{Im } h_n \Longrightarrow \exists \ x_n\) such that \(\displaystyle h_n ( x_n ) = y_n \) Then ... \(\displaystyle \exists \ y \in N\) such that \(\displaystyle y = y_1 + y_2 + y_3 + \ ... \ ... \ + y_n\) ... ... since \(\displaystyle N\) is a module ...

and then \(\displaystyle y \in \text{Im } h_1 + \in \text{Im } h_2 + \in \text{Im } h_3 + \ ... \ ... \ + \text{Im } h_n\)Now \(\displaystyle y = y_1 + y_2 + y_3 + \ ... \ ... \ + y_n \Longrightarrow y = h_1(x_1) + h_2(x_2) + h_3(x_3) + \ ... \ ... \ + h_n(x_n) = \sum_\Delta h_\alpha ( x_\alpha ) \)

That is \(\displaystyle y = f(x)\) where \(\displaystyle x = ( x_1, x_2, x_3, \ ... \ ... \ , x_n ) = (x_\alpha)_\Delta\) ...

So \(\displaystyle f\) is an epimorphism ...Is that basically correct ...

Peter

 

[steenis]

It is basically correct, but change:

Then ... \(\displaystyle \exists \ y \in N\) such that \(\displaystyle y = y_1 + y_2 + y_3 + \ ... \ ... \ + y_n\) ... ... since \(\displaystyle N\) is a module ...

and then \(\displaystyle y \in \text{Im } h_1 + \in \text{Im } h_2 + \in \text{Im } h_3 + \ ... \ ... \ + \text{Im } h_n\)

Now \(\displaystyle y = y_1 + y_2 + y_3 + \ ... \ ... \ + y_n \Longrightarrow y = h_1(x_1) + h_2(x_2) + h_3(x_3) + \ ... \ ... \ + h_n(x_n) = \sum_\Delta h_\alpha ( x_\alpha ) \)

into:
Define $y = y_1 + y_2 + \cdots + y_n$ then $y \in N$, because N is a module.

And $y = h_1(x_1) + h_2(x_2) + \cdots + h_n(x_n) = \sum_\Delta h_\alpha ( x_\alpha ) \in \Sigma_\Delta \text{ I am } h_\alpha = N$