# Modules Generated by Sets of Submodules .... .... Bland Problem 2, Problem Set 4.1 .... ....

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In summary: I will try and work on Problem 2 tomorrow morning ... In summary, Peter found that there is a $y \in N$ such that $f(x) \neq 0$ and $f \circ \phi \neq 0$.
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I am reading Paul E. Bland's book "Rings and Their Modules" ...

Currently I am focused on Section 4.1 Generating and Cogenerating Classes ... ...

I need some help in order to make a meaningful start on Problem 2, Problem Set 4.1 ...

Problem 2, Problem Set 4.1 reads as follows:( *** NOTE ... due to some problem I cannot upload scans/images at the moment ... so until the problem is resolved MHB members will need to have access to Bland's book ... Oh! image appeared on my screen at end of post ... )Can someone please help me to make a meaningful start on Problem 2 ...

Peter

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$\Rightarrow )$
Let $\phi:M^{(\Delta)} = \bigoplus_\Delta M_\alpha \to N$ be the “generating epimorphism”, $M_\alpha = M$.

$i_\alpha: M_\alpha = M \to \bigoplus_\Delta M_\alpha$ are the canonical inclusions for all $\alpha \in \Delta$.

Define $\phi_\alpha = \phi \circ i_\alpha$.

$f:N \to N’$ is given, $f \neq 0$.

1) make a diagram
2) prove that $f \circ \phi \neq 0$

steenis said:
$\Rightarrow )$
Let $\phi:M^{(\Delta)} = \bigoplus_\Delta M_\alpha \to N$ be the “generating epimorphism”, $M_\alpha = M$.

$i_\alpha: M_\alpha = M \to \bigoplus_\Delta M_\alpha$ are the canonical inclusions for all $\alpha \in \Delta$.

Define $\phi_\alpha = \phi \circ i_\alpha$.

$f:N \to N’$ is given, $f \neq 0$.

1) make a diagram
2) prove that $f \circ \phi \neq 0$
Thanks Steenis ...

Late now in southern Tasmania (edge of the world Strong winds ... 115 km per hour roaring round my old house ... )

Will work on problem ... based on your advice ... tomorrow morning...

Thanks again ...

Peter

steenis said:
$\Rightarrow )$
Let $\phi:M^{(\Delta)} = \bigoplus_\Delta M_\alpha \to N$ be the “generating epimorphism”, $M_\alpha = M$.

$i_\alpha: M_\alpha = M \to \bigoplus_\Delta M_\alpha$ are the canonical inclusions for all $\alpha \in \Delta$.

Define $\phi_\alpha = \phi \circ i_\alpha$.

$f:N \to N’$ is given, $f \neq 0$.

1) make a diagram
2) prove that $f \circ \phi \neq 0$
Thanks for the advice ...!We are trying to show that

$$\displaystyle M \text{ generates } N \Longrightarrow$$ for each non-zero R-linear mapping $$\displaystyle f \ : \ N \rightarrow N'$$ there is an R-linear mapping $$\displaystyle h \ : \ M \rightarrow N$$ such that $$\displaystyle f \circ h \ne 0$$ ... ...
Now $$\displaystyle M$$ generates $$\displaystyle N \Longrightarrow \exists$$ an epimorphism $$\displaystyle \phi \ : \ M^{(\Delta)} = \bigoplus_\Delta M_\alpha \to N$$ ...

... and we define $$\displaystyle i_\alpha: M_\alpha = M \to \bigoplus_\Delta M_\alpha$$ ...

... and further define $$\displaystyle \phi_\alpha = \phi \circ i_\alpha$$ ...

and we are given $$\displaystyle f \ : \ N \rightarrow N'$$ ...A diagram of this is shown below:View attachment 8146
Now ... you asked me to show that $$\displaystyle f \circ \phi \ne 0$$ ...

That is ... it is not the case that $$\displaystyle f \circ \phi (x) = f( \phi (x) ) = 0$$ for all $$\displaystyle x \in \bigoplus_\Delta M_\alpha$$ ...

(best to find a particular x such that $$\displaystyle f \circ \phi (x) = y \ne 0$$ ... but how ...? maybe $$\displaystyle 1$$ maps to $$\displaystyle 1$$ ...?)

But $$\displaystyle \phi$$ is an epimorphism and so maps to all of $$\displaystyle N$$, and then $$\displaystyle f \ne 0$$ ... so we can conclude that $$\displaystyle f \circ \phi \ne 0$$ ... ... but ... ... is this a valid argument ... ? ... suspect not ..Then how do we progress from here ...

Can you help further ... ?

Peter
***NOTE***

... ... it seems to me that $$\displaystyle \phi$$ might be our required R-linear map such that $$\displaystyle f \circ h = f \circ \phi \ne 0$$ ... $$\displaystyle \phi$$ is indeed R-linear ... BUT ... $$\displaystyle \phi$$ maps from $$\displaystyle M^{(\Delta)}$$ to $$\displaystyle N$$ and we are looking for a mapping $$\displaystyle h$$ that maps from $$\displaystyle M$$ to $$\displaystyle N$$ ...Hmm ... Is $$\displaystyle \phi_\alpha$$ the map we want ... do we need to show $$\displaystyle f \circ \phi_\alpha \ne 0$$ and then take $$\displaystyle \phi_\alpha = h$$ ... ... but then \{ \phi_\alpha \} is not one mapping but a family of mappings ...

Can you help ... ?

Peter

Last edited:
You were almost there: $f \neq 0$, so there is a $y \in N$ with $f(x) \neq 0$.
$\phi$ is an epimorpism, so there is a $y \in \bigoplus_\Delta M_\alpha$, such that $\phi (y) = x$,
it follows that $\phi (f(x)) \neq 0$ and $f \circ \phi \neq 0$.

3) use magical prop.2.1.5 to expand $\phi$ into a sum (see a former thread of you)
4) prove there is an $\alpha \in \Delta$ such that $f \circ \phi_\alpha \neq 0$
5) conclusion

steenis said:
You were almost there: $f \neq 0$, so there is a $y \in N$ with $f(x) \neq 0$.
$\phi$ is an epimorpism, so there is a $y \in \bigoplus_\Delta M_\alpha$, such that $\phi (y) = x$,
it follows that $\phi (f(x)) \neq 0$ and $f \circ \phi \neq 0$.

3) use magical prop.2.1.5 to expand $\phi$ into a sum (see a former thread of you)
4) prove there is an $\alpha \in \Delta$ such that $f \circ \phi_\alpha \neq 0$
5) conclusion
Thanks steenis ...

Peter

steenis said:
You were almost there: $f \neq 0$, so there is a $y \in N$ with $f(x) \neq 0$.
$\phi$ is an epimorpism, so there is a $y \in \bigoplus_\Delta M_\alpha$, such that $\phi (y) = x$,
it follows that $\phi (f(x)) \neq 0$ and $f \circ \phi \neq 0$.

3) use magical prop.2.1.5 to expand $\phi$ into a sum (see a former thread of you)
4) prove there is an $\alpha \in \Delta$ such that $f \circ \phi_\alpha \neq 0$
5) conclusion
Thanks again steenis ...

... But ... just a clarification ... $f \neq 0$, so there is a $y \in N$ with $f(x) \neq 0$. ... ... "Did you mean: " ... ... $f \neq 0$, so there is a $x \in N$ with $f(x) \neq 0$. ... ... "

Peter

You write:

" ... ...

Yes, of course, I am sorry.

and also $f(\phi (y)) \neq 0$, sorry again. I think I was still sleeping ...

steenis said:
Yes, of course, I am sorry.

and also $f(\phi (y)) \neq 0$, sorry again. I think I was still sleeping ...
Thanks ... OK now ...

Peter

steenis said:
Yes, of course, I am sorry.

and also $f(\phi (y)) \neq 0$, sorry again. I think I was still sleeping ...
Now consider $$\displaystyle \phi \ : \ M^{(\Delta)} = \bigoplus_\Delta M_\alpha \to N$$Let $$\displaystyle ( x_\alpha ) \in \bigoplus_\Delta M_\alpha$$ ...Then by Proposition 2.1.5 we have ...

$$\displaystyle \phi ( ( x_\alpha ) ) = \sum_\Delta \phi_\alpha ( x_\alpha )$$ ... ... where the \phi_\alpha are R-linear maps ... ...Now since $$\displaystyle f \circ \phi \ne 0$$ we have $$\displaystyle f \circ \sum_\Delta \phi_\alpha \ne 0$$ ... ... Now ... $$\displaystyle f \circ \sum_\Delta \phi_\alpha \ne 0$$$$\displaystyle \Longrightarrow$$ for at least one $$\displaystyle \alpha$$ we have $$\displaystyle f \circ \phi_\alpha \ne 0$$ ... ... ... (is this correct?)Now fix/choose one of these $$\displaystyle \alpha$$ so that $$\displaystyle f \circ \phi_\alpha \ne 0$$ ...Then put $$\displaystyle h = \phi_\alpha$$Then $$\displaystyle \exists \ h$$ so that $$\displaystyle f \circ h \ne 0$$ ...Is that correct ...

Peter

All correct, well done.

Peter, there is something wrong with the site of MHB. I will be bach soon

I myself did something wrong.

For the converse you have to construct an R-epimorhism:
$$\phi:M^{(\Delta)} = \bigoplus_\Delta M_\alpha \to N$$
$(M_\alpha = M)$.
This means that you have to find the index-set $\Delta$. I did not find it myself so I will give it away: $\Delta = \text{Hom } (M,N)$. Thus, you use $\text{Hom } (M,N)$ as an index set.

Now use prop.2.1.5 to construct $\phi$, you have done that before. Make diagrams (don’t post them, I trust you can make them).
Last step is to prove that $\phi$ is surjective.
Suppose that $\phi$ is not surjective, what can you tell about $N/\text{im } \phi$ ?
Finally use the hypothesis.

steenis said:
I myself did something wrong.

For the converse you have to construct an R-epimorhism:
$$\phi:M^{(\Delta)} = \bigoplus_\Delta M_\alpha \to N$$
$(M_\alpha = M)$.
This means that you have to find the index-set $\Delta$. I did not find it myself so I will give it away: $\Delta = \text{Hom } (M,N)$. Thus, you use $\text{Hom } (M,N)$ as an index set.

Now use prop.2.1.5 to construct $\phi$, you have done that before. Make diagrams (don’t post them, I trust you can make them).
Last step is to prove that $\phi$ is surjective.
Suppose that $\phi$ is not surjective, what can you tell about $N/\text{im } \phi$ ?
Finally use the hypothesis.
Hi steenis ... despite your help (thank you) ... I have not been able to make much progress on the converse ... see below ...
Converse:If for each non-zero R-linear mapping $$\displaystyle f \ : \ N \to N' \ \ \exists$$ an R-linear mapping $$\displaystyle h$$ such that $$\displaystyle f \circ h \ne 0$$ ...

then $$\displaystyle M$$ generates $$\displaystyle N$$ ...

... that is ... $$\displaystyle \exists$$ an epimorphism $$\displaystyle \phi \ : \ M^{ ( \Delta ) } = \bigoplus_\Delta M_\alpha$$ for some set $$\displaystyle \Delta$$ ...Now ... take $$\displaystyle \Delta = \text{ Hom } (M, N)$$ ... ... I am not sure why ... can you elaborate ... ?
So ... let $$\displaystyle i_\alpha \ : \ M_\alpha \to \bigoplus_\Delta M_\alpha$$

and $$\displaystyle \phi_\alpha \ : \ M_\alpha \to N$$ ...

... and then define $$\displaystyle \phi_\alpha = \phi \circ i_\alpha$$ ...Now consider a non-zero R-linear mapping $$\displaystyle f \ : \ N \to N'$$ ...Now we have by our hypothesis that there exists an R-linear mapping $$\displaystyle h \ : \ M \to N$$ such that $$\displaystyle f \circ h \ne 0$$ ... ...... BUT ... where to from here?

Can we take $$\displaystyle h$$ to be equal to $$\displaystyle \phi_\alpha$$ ... hmm ... I don't think we can ...How to proceed? ... can you help further ...Further ... you write: " ... Suppose that $\phi$ is not surjective, what can you tell about $N/\text{im } \phi$ ? ... ... "

Not sure what you can tell about $N/\text{im } \phi$ ... ... can you elaborate ...

Peter

Proc.2.1.5 does this:

Given an index set $\Delta$

Given a family of R-modules $\{M_\alpha | \alpha \in \Delta\}$

Given an R-module $N$

Given a family of R-maps $\{\phi_\alpha : M_\alpha \to N\}$

Then you can construct a unique R-map

$\phi:\bigoplus_\Delta M_\alpha \to N$ satisfying

$\phi \circ i_\alpha = \phi_\alpha$, where

$i_\alpha : M_\alpha \to \bigoplus_\Delta M_\alpha$ are the canonical injections (Make a diagram)

The constructed R-map $\phi$ is defined on $\bigoplus_\Delta M_\alpha$ in this way:

$\phi((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } \phi_\alpha (x_\alpha)$ for $((x_\alpha)_\Delta) \in \bigoplus_\Delta M_\alpha$

Now you repeat this same procedure given the index set $\Delta = \text{Hom }(M, N)$
(We elobarate on $\text{Hom }(M, N)$ later, if necessary)
(we define the $\phi_\alpha$ later)

In the second procedure: $M_\alpha = M$, of course

steenis said:
Proc.2.1.5 does this:

Given an index set $\Delta$

Given a family of R-modules $\{M_\alpha | \alpha \in \Delta\}$

Given an R-module $N$

Given a family of R-maps $\{\phi_\alpha : M_\alpha \to N\}$

Then you can construct a unique R-map

$\phi:\bigoplus_\Delta M_\alpha \to N$ satisfying

$\phi \circ i_\alpha = \phi_\alpha$, where

$i_\alpha : M_\alpha \to \bigoplus_\Delta M_\alpha$ are the canonical injections (Make a diagram)

The constructed R-map $\phi$ is defined on $\bigoplus_\Delta M_\alpha$ in this way:

$\phi((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } \phi_\alpha (x_\alpha)$ for $((x_\alpha)_\Delta) \in \bigoplus_\Delta M_\alpha$

Now you repeat this same procedure given the index set $\Delta = \text{Hom }(M, N)$
(We elobarate on $\text{Hom }(M, N)$ later, if necessary)
(we define the $\phi_\alpha$ later)
Thanks for the help, Steenis ...

Certainly over the last few problems, you have helped me to better understand Proposition 2.15 (and hence the nature of the other similar propositions in Section 2.1) ... now I think I have an idea of its nature ...

Just a couple of points ... you write:

" ... ... (We elobarate on $\text{Hom }(M, N)$ later, if necessary) ... ... "It does seem a strange index set to me ... being a set of mappings ... but then I guess as an index set only the correct cardinality matters ...

You also write:

" ... ... (we define the $\phi_\alpha$ later) ... ... " But ... haven't we defined $\phi_\alpha$ when we write: ... ... $\phi \circ i_\alpha = \phi_\alpha$ ...?Thanks again for all your help ...

Peter

steenis said:
In the second procedure: $M_\alpha = M$, of course
Hi Steenis ... Thanks to you clarifying the use of Proposition 2.1.5 I can now see that if we define $$\displaystyle \Delta = \text{ Hom } ( M, N )$$ and then define/assume a family of R-modules $$\displaystyle M_\alpha$$, an R-module $$\displaystyle N$$ and a family of R-linear mappings, $$\displaystyle \phi_\alpha$$ then Proposition 2.1.5 guarantees the existence of a unique R-linear mapping:

$$\displaystyle \phi \ : \ \bigoplus_{ \text{ Hom } ( M, N ) } M_\alpha \to N$$

satisfying:

$$\displaystyle \phi_\alpha = \phi \circ i_\alpha$$ So ... we have a map $$\displaystyle \phi$$ from $$\displaystyle \bigoplus_{ \text{ Hom } ( M, N ) } M_\alpha$$ to $$\displaystyle N$$ (where $$\displaystyle M_\alpha = M$$ for all $$\displaystyle \alpha \in \text{ Hom } ( M, N )$$ ...

Now ... we need to show that the R-linear map $$\displaystyle \phi$$ is surjective ... but how ...

... hmmm ... been thinking of your hint to assume $$\displaystyle \phi$$ not surjective and then consider nature of $$\displaystyle N/ \text{ I am } \phi$$... but can't make any headway about what we know about $$\displaystyle N/ \text{ I am } \phi$$ in the case where $$\displaystyle \phi$$ is not surjective ...Hmmm ... have not yet used the hypothesis regarding $$\displaystyle f$$ and $$\displaystyle h$$ yet ... but ... how do we employ it to show that $$\displaystyle \phi$$ is surjective ... ?
Sorry about lack of real progress ... but can you help further ...

Peter

Last edited:
I am disappointed that you did not answer my question:

"Now you repeat this same procedure given the index set $\Delta = \text{Hom }(M, N)$"

As long as we do not know the wanted R-map $\phi$, we cannot go further. So please construct $\phi$, and while doing that make a suggestion what the $\phi_\alpha$ are. Make a good choice for the index-variable $\alpha$. Of course make diagrams.

steenis said:
I am disappointed that you did not answer my question:

"Now you repeat this same procedure given the index set $\Delta = \text{Hom }(M, N)$"

As long as we do not know the wanted R-map $\phi$, we cannot go further. So please construct $\phi$, and while doing that make a suggestion what the $\phi_\alpha$ are. Make a good choice for the index-variable $\alpha$. Of course make diagrams.
But, Hugo, repeating the procedure with $\Delta = \text{Hom }(M, N)$ just gives (as I stated) the mapping

$$\displaystyle \phi \ : \ \bigoplus_{ \text{ Hom } ( M, N ) } M_\alpha \to N$$

where the copies of M are counted by the number of homomorphisms in \text{Hom }(M, N) ...But then, from your comment I am assuming that I am missing something ...

Peter

Yes, you are missing the construction of $\phi$. Please read my post #14

Now you repeat this same procedure given the index set $\Delta = \text{Hom }(M, N)$

Decide what the $\phi_\alpha$ are.

steenis said:
I am disappointed that you did not answer my question:

"Now you repeat this same procedure given the index set $\Delta = \text{Hom }(M, N)$"

As long as we do not know the wanted R-map $\phi$, we cannot go further. So please construct $\phi$, and while doing that make a suggestion what the $\phi_\alpha$ are. Make a good choice for the index-variable $\alpha$. Of course make diagrams.
hi steenis ...

You write:

" ... ... make a suggestion what the $\phi_\alpha$ are. ... ... "I am thinking that the $\phi_\alpha$ are the h of the hypothesis ... so that $$\displaystyle f \circ h = f \circ \phi_\alpha \ne 0$$ ... and that somehow we use this to show that $$\displaystyle \phi$$ is a surjection ...

Peter

steenis said:
Yes, you are missing the construction of $\phi$. Please read my post #14

Now you repeat this same procedure given the index set $\Delta = \text{Hom }(M, N)$

Decide what the $\phi_\alpha$ are.
I can see that the map $$\displaystyle \phi \ : \ \bigoplus_{ \text{ Hom } ( M, N ) } M_\alpha \to N$$

is defined by

$\phi((x_\alpha)_{ \text{ Hom } ( M, N ) } ) = \Sigma_{ \text{ Hom } ( M, N ) } \text{ } \phi_\alpha (x_\alpha)$ for $((x_\alpha)_{ \text{ Hom } ( M, N ) }) \in \bigoplus_\Delta M_\alpha$But ... I am thinking of the index set as a counter ... but maybe that is limiting or wrong ... in fact I am puzzled as to how a set of homomorphisms can act as a counter similar to, say, the natural numbers ... can you explain how a set of homomorphisms acts as an index ...Peter

steenis said:
Proc.2.1.5 does this:

Given an index set $\Delta$

Given a family of R-modules $\{M_\alpha | \alpha \in \Delta\}$

Given an R-module $N$

Given a family of R-maps $\{\phi_\alpha : M_\alpha \to N\}$

Then you can construct a unique R-map

$\phi:\bigoplus_\Delta M_\alpha \to N$ satisfying

$\phi \circ i_\alpha = \phi_\alpha$, where

$i_\alpha : M_\alpha \to \bigoplus_\Delta M_\alpha$ are the canonical injections (Make a diagram)

The constructed R-map $\phi$ is defined on $\bigoplus_\Delta M_\alpha$ in this way:

$\phi((x_\alpha)_\Delta) = \Sigma_\Delta \text{ } \phi_\alpha (x_\alpha)$ for $((x_\alpha)_\Delta) \in \bigoplus_\Delta M_\alpha$

Now you repeat this same procedure given the index set $\Delta = \text{Hom }(M, N)$
(We elobarate on $\text{Hom }(M, N)$ later, if necessary)
(we define the $\phi_\alpha$ later)
I do not know why you refuse to answer my question, I was trying to make things clear to you. Well I will answer it and do the procedure:Given the index set $\Delta = \text{Hom }(M, N)$

Instead of the index-variable $\alpha \in \Delta$, I will use the index-variable $f \in \text{Hom }(M, N)$

Given a family of R-modules $\{M_f = M | f \in \text{Hom }(M, N)\}$

Given an R-module $N$Given a family of R-maps $\{\phi_f : M_f \to N\}$
NO these $\phi_f$ are not given.
I suggest $\phi_f = f$ for $f \in \text{Hom }(M, N)$.

Then you can construct a unique R-map

$\phi:\bigoplus_{\text{Hom }(M, N)} M_f \to N$ satisfying

$\phi \circ i_f = \phi_f = f$, where

$i_f : M_f \to \bigoplus_{\text{Hom }(M, N)} M_f$ are the canonical injections

The constructed R-map $\phi$ is defined on $\bigoplus_{\text{Hom }(M, N)} M_f$ in this way:

$\phi((x_f)_{\text{Hom }(M, N)}) = \Sigma_{\text{Hom }(M, N)} \text{ } f(x_f)$ for $((x_f)_{\text{Hom }(M, N)}) \in \bigoplus_{\text{Hom }(M, N)} M_f$

Now you should have the diagram:

\begin{tikzpicture}[>=stealth]
\node (A) at (1,0) {$M_f = M$};
\node (B) at (1,-6) {$\bigoplus_{Hom (M, N)} M_f$};
\node (C) at (6,-6) {$N$};
\draw[->] (A) -- node
{$i_f$} (B);
\draw[->] (A) -- node
{$f$} (C);
\draw[->] (B) -- node [below] {$\phi$} (C);
\end{tikzpicture}​

steenis said:
I do not know why you refuse to answer my question, I was trying to make things clear to you. Well I will answer it and do the procedure:Given the index set $\Delta = \text{Hom }(M, N)$

Instead of the index-variable $\alpha \in \Delta$, I will use the index-variable $f \in \text{Hom }(M, N)$

Given a family of R-modules $\{M_f = M | f \in \text{Hom }(M, N)\}$

Given an R-module $N$Given a family of R-maps $\{\phi_f : M_f \to N\}$
NO these $\phi_f$ are not given.
I suggest $\phi_f = f$ for $f \in \text{Hom }(M, N)$.

Then you can construct a unique R-map

$\phi:\bigoplus_{\text{Hom }(M, N)} M_f \to N$ satisfying

$\phi \circ i_f = \phi_f = f$, where

$i_f : M_f \to \bigoplus_{\text{Hom }(M, N)} M_f$ are the canonical injections

The constructed R-map $\phi$ is defined on $\bigoplus_{\text{Hom }(M, N)} M_f$ in this way:

$\phi((x_f)_{\text{Hom }(M, N)}) = \Sigma_{\text{Hom }(M, N)} \text{ } f(x_f)$ for $((x_f)_{\text{Hom }(M, N)}) \in \bigoplus_{\text{Hom }(M, N)} M_f$

Now you should have the diagram:

\begin{tikzpicture}[>=stealth]
\node (A) at (1,0) {$M_f = M$};
\node (B) at (1,-6) {$\bigoplus_{Hom (M, N)} M_f$};
\node (C) at (6,-6) {$N$};
\draw[->] (A) -- node
{$i_f$} (B);
\draw[->] (A) -- node
{$f$} (C);
\draw[->] (B) -- node [below] {$\phi$} (C);
\end{tikzpicture}​
Thanks steenis ...

Will study it now and tomorrow morning ...

Thank you for your patience ...

Peter​

steenis said:
I do not know why you refuse to answer my question, I was trying to make things clear to you. Well I will answer it and do the procedure:Given the index set $\Delta = \text{Hom }(M, N)$

Instead of the index-variable $\alpha \in \Delta$, I will use the index-variable $f \in \text{Hom }(M, N)$

Given a family of R-modules $\{M_f = M | f \in \text{Hom }(M, N)\}$

Given an R-module $N$Given a family of R-maps $\{\phi_f : M_f \to N\}$
NO these $\phi_f$ are not given.
I suggest $\phi_f = f$ for $f \in \text{Hom }(M, N)$.

Then you can construct a unique R-map

$\phi:\bigoplus_{\text{Hom }(M, N)} M_f \to N$ satisfying

$\phi \circ i_f = \phi_f = f$, where

$i_f : M_f \to \bigoplus_{\text{Hom }(M, N)} M_f$ are the canonical injections

The constructed R-map $\phi$ is defined on $\bigoplus_{\text{Hom }(M, N)} M_f$ in this way:

$\phi((x_f)_{\text{Hom }(M, N)}) = \Sigma_{\text{Hom }(M, N)} \text{ } f(x_f)$ for $((x_f)_{\text{Hom }(M, N)}) \in \bigoplus_{\text{Hom }(M, N)} M_f$

Now you should have the diagram:

\begin{tikzpicture}[>=stealth]
\node (A) at (1,0) {$M_f = M$};
\node (B) at (1,-6) {$\bigoplus_{Hom (M, N)} M_f$};
\node (C) at (6,-6) {$N$};
\draw[->] (A) -- node
{$i_f$} (B);
\draw[->] (A) -- node
{$f$} (C);
\draw[->] (B) -- node [below] {$\phi$} (C);
\end{tikzpicture}​

Just a thought Hugo ... $$\displaystyle f$$ is the notation for a mapping $$\displaystyle f \ : \ N \to N'$$ in the hypothesis of the converse ...

Won't this cause confusion as we proceed ...

But anyway ... for now focusing on your post and its meaning

Peter​

We will use another notation.

steenis said:
We will use another notation.
OK Hugo ... thanks ...

I now believe I have understood your post ... at least superficially anyway ...

But ... still cannot see the way forward to prove that $$\displaystyle \phi$$ is surjective ...

I may need some more help ... :( ... apologies ...

Peter

The only thing I want to know if you understand the procedure of the construction of $\phi$. If not, go back to post #14, study the procedure of prop.2.1.5. Then you repeat the procedure for $\Delta = \text{Hom }(M, N)$ without peeking. Choose an appropriate index-variable $\alpha$ and appropriate R-map $\phi_\alpha$. Only if you understand it thoroughly, we wil continue and prove that $\phi$ is surjective.

steenis said:
The only thing I want to know if you understand the procedure of the construction of $\phi$. If not, go back to post #14, study the procedure of prop.2.1.5. Then you repeat the procedure for $\Delta = \text{Hom }(M, N)$ without peeking. Choose an appropriate index-variable $\alpha$ and appropriate R-map $\phi_\alpha$. Only if you understand it thoroughly, we wil continue and prove that $\phi$ is surjective.
I believe I understand the procedure ... but I am ging back once again to your post #14 to study it once again ...

Peter

steenis said:
As I see it ... we are given a family of R-modules $\{M_\alpha \mid \alpha \in \Delta\}$ ... then we can take any R-module N and any family of R-linear mappings $\{\phi_\alpha \mid M_\alpha \to N\}$ and then be assured that there exists a uniqe R-linear mapping $\phi:\bigoplus_\Delta M_\alpha \to N$ satisfying

$\phi \circ i_\alpha = \phi_\alpha$, where

$i_\alpha : M_\alpha \to \bigoplus_\Delta M_\alpha$ are the canonical injections ...

I believe that is the correct logic for constructing $$\displaystyle \phi$$ ...Now to repeat the process with $$\displaystyle \Delta = \text{ Hom } ( M, N )$$ it seems to me we just replace $$\displaystyle \Delta$$ with $$\displaystyle \text{ Hom } ( M, N )$$ ... this is where my understanding may be a bit shallow since replacing the index set doesn't seem to me to make much difference to the proof ... so I may be missing something of significance ...

Peter

That is a correct interpretation. But now you are applying this proposition, sot you have to fill in the "unknowns", as you well know.

$\text{Hom }(M, N)$ is a very special index-set. Furthermore, if you follow the procedure you see that the R-maps $\phi_\alpha$ are not defined. So you have to define them.

steenis said:
That is a correct interpretation. But now you are applying this proposition, sot you have to fill in the "unknowns", as you well know.

$\text{Hom }(M, N)$ is a very special index-set. Furthermore, if you follow the procedure you see that the R-maps $\phi_\alpha$ are not defined. So you have to define them.
When you defined the $$\displaystyle \phi_\alpha$$ as $$\displaystyle \phi_\alpha = s$$ where $$\displaystyle s \in \text{ Hom } ( M, N )$$ ... ... was that a sufficiently precise definition ... or do we need to define the mapping between elements of the domain and image sets ...

I understand that ..." ... $\text{Hom }(M, N)$ is a very special index-set. ... " but how it affects the logic of the proof apart form specifying the index I don't see ...

Still thinking ...

Peter

No, because we use $\text{Hom }(M, N)$ as an index set.

If you have a theorem and you want to apply this theorem what do you do ?

steenis said:
No, because we use $\text{Hom }(M, N)$ as an index set.

If you have a theorem and you want to apply this theorem what do you do ?
You make sure that the conditions in the hypothesis are met ,,,

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