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

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In summary: R$are very special submodules of$M$that “span”$M$. You could ask yourself: can$M$be spanned by more general submodules of$M$? For instance,$M = A + B$, with$A, B \leq M$, when for each$m \in M$there is a$a \in A$and a$b \in B$such that$m = a + b$.In this light we can say that if$M = \Sigma N_\alpha$and$N_\alpha \leq M$then$M$is spanned by the submodules$N_\alpha \le
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.

steenis said:
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

It is basically correct, but change:

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

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