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

[Math Amateur]

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

 

[steenis]

$\Rightarrow )$
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 $f \circ \phi \neq 0$
What is your next step?

 

[Math Amateur]

$\Rightarrow )$
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 $f \circ \phi \neq 0$
What is your next step?

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

 

[Math Amateur]

$\Rightarrow )$
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 $f \circ \phi \neq 0$
What is your next step?

Thanks for the advice ...!We are trying to show that

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

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

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

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

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

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

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

Can you help ... ?

Peter

 

[steenis]

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

 

[Math Amateur]

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 ...

Reflecting on your advice now ...

Peter

 

[Math Amateur]

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:

" ... ...

 

[steenis]

Yes, of course, I am sorry.

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

 

[Math Amateur]

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

 

[Math Amateur]

Yes, of course, I am sorry.

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

Now consider $$\phi \ : \ M^{(\Delta)} = \bigoplus_\Delta M_\alpha \to N$$Let $$( x_\alpha ) \in \bigoplus_\Delta M_\alpha$$ ...Then by Proposition 2.1.5 we have ...

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

Peter

 

[steenis]

All correct, well done.

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

 

[steenis]

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.

 

[Math Amateur]

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 $$f \ : \ N \to N' \ \ \exists$$ an R-linear mapping $$h$$ such that $$f \circ h \ne 0$$ ...

then $$M$$ generates $$N$$ ...

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

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

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

Can we take $$h$$ to be equal to $$\phi_\alpha$$ ... 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

 

[steenis]

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)

 

[steenis]

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

 

[Math Amateur]

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

 

[Math Amateur]

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 $$\Delta = \text{ Hom } ( M, N )$$ and then define/assume a family of R-modules $$M_\alpha$$, an R-module $$N$$ and a family of R-linear mappings, $$\phi_\alpha$$ then Proposition 2.1.5 guarantees the existence of a unique R-linear mapping:

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

satisfying:

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

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

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

Peter

 

[steenis]

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.

 

[Math Amateur]

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

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

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

 

[steenis]

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.

 

[Math Amateur]

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 $$f \circ h = f \circ \phi_\alpha \ne 0$$ ... and that somehow we use this to show that $$\phi$$ is a surjection ...

Peter

 

[Math Amateur]

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

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]

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}​

 

[Math Amateur]

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 ...

Have read your post ... but do not follow it yet ...

Will study it now and tomorrow morning ...

Thank you for your patience ...

Peter​

 

[Math Amateur]

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 ... $$f$$ is the notation for a mapping $$f \ : \ N \to N'$$ 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​

 

[steenis]

We will use another notation.

 

[Math Amateur]

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 $$\phi$$ is surjective ...

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

Peter

 

[steenis]

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.

 

[Math Amateur]

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]

Good, take your time

 

[Math Amateur]

Good, take your time

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 $$\phi$$ ...Now to repeat the process with $$\Delta = \text{ Hom } ( M, N )$$ it seems to me we just replace $$\Delta$$ with $$\text{ Hom } ( M, N )$$ ... 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

 

[steenis]

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.

 

[Math Amateur]

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 $$\phi_\alpha$$ as $$\phi_\alpha = s$$ where $$s \in \text{ Hom } ( M, N ) $$ ... ... 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

 

[steenis]

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 ?

 

[Math Amateur]

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 ,,,

 

[steenis]

So do that, apply prop.2.1.5 to our situation

 

[Math Amateur]

So do that, apply prop.2.1.5 to our situation

I suspect that will require some forward looking and creative definitions ... but I'll try ..

I am still very unsure how using $$\text{ Hom } ( M, N )$$ as an index affect things or enters the proof in any significant way ...

Must go to sleep now ... it is after midnight ...

Will continue in the morning ...

Peter*** EDIT ***

Do you feel able to give any further hints on $$N/ \text{ I am } \phi $$

 

[steenis]

Suppose $\phi$ is not surjective, what can you say about $\text{ I am } \phi$ and $N/\text{ I am }\phi$ ?

 

[Math Amateur]

Suppose $\phi$ is not surjective, what can you say about $\text{ I am } \phi$ and $N/\text{ I am }\phi$ ?

If \phi is not surjective then $$\text{ I am } \phi \ne N$$ and indeed $$\text{ I am } \phi \subset N$$ ...

But $N/\text{ I am }\phi$ ... ... what can we say ...?

PeterEDIT ... are you thinking of Correspondence Theorem ..?

 

[steenis]

If $\text{ I am }\phi \neq N$ then $N/\text{ I am }\phi \neq 0$. What can you say about the canonical projection $v:N \to N/\text{ I am }\phi$ ?

 

[Math Amateur]

If $\text{ I am }\phi \neq N$ then $N/\text{ I am }\phi \neq 0$. What can you say about the canonical projection $v:N \to N/\text{ I am }\phi$ ?

You can say that $$v \ne 0$$ ... that is v is not equal to the zero map ...

 

[steenis]

what next ?

 

[Math Amateur]

what next ?

Not sure what else follows ...

But it is nearly 2am ...

Can you give me a hint to continue with tomorrow morning ,,.

Falling asleep ...

Peter

 

[steenis]

Apply the hypothesis ...

 

[steenis]

In mathematics, an index set is a set whose members label (or index) members of another set. For instance, if the elements of a set $A$ may be indexed or labeled by means of a set $J$, then $J$ is an index set. The indexing consists of a surjective function from $J$ onto $A$ and the indexed collection is typically called an (indexed) family, often written as $(A_j)_{j\in J}$, or $\{x_j \in A\}_{j\in J}$. [See: Munkres – Topology (2000) p.36]
As far as I know, every set can be used as an index-set, as long as it is a set.

 

[Math Amateur]

Apply the hypothesis ...

In an attempt to apply the hypothesis in a way that takes account of the hints you have given ... in the hope that something emerges ... like a contradiction or a chance to apply a contrapositive ... Put $$f = v$$ where $$v \ : \ N \to N / \text{ I am } \phi$$ is such that $$v(x) = \overline{x} + \text{ I am } \phi$$ ...Also put $$h = \phi_\alpha$$ ...Then we have the situation in Figure 2 below ...https://www.physicsforums.com/attachments/8164

BUT ... the situation with the definitions I've made is perplexing ... if $$\phi$$ is not an epimorphism ... then $$f = v \ne 0$$ as required by the converse hypothesis ... and I suspect (but not completely sure... ) that $$f \circ h = v \circ \phi_\alpha$$ is also not equal to zero ... but then the conditions of the converse hypothesis are fulfilled ... assuming that (contrary to the implications of the problem statement) $$\phi$$ is NOT an epimorphism ... ? ... something wrong ... possibly $$f \circ h = v \circ \phi_\alpha$$ is NOT non-zero ... Can you clarify ...

Are you able to hint or show me how to make a better set of definitions that still take account of your hints regarding $$\text{ I am } \phi$$ and $$N / \text{ I am } \phi$$ ...Peter
***EDIT***

Attempt to investigate whether $$f \circ h \ne 0$$We have ...

$$f \circ h ( x_\alpha ) = v \circ \phi_\alpha ( x_\alpha )$$$$= v \circ \phi \circ i_\alpha ( x_\alpha ) $$$$= v \circ \phi ( 0 ... ... x_\alpha ... ... 0 ) $$ $$= v ( \sum_{ \text{ Hom } ( M, N ) } ( x_\beta )$$ where $$x_\beta = 0$$ when $$\beta \ne \alpha$$ and $$x_\beta = x_\alpha$$ when $$\beta = \alpha$$$$= v \circ \phi_\alpha ( x_\alpha )$$ since $$\phi_\alpha (0) = 0$$ $$= \overline{ \phi_\alpha ( x_\alpha ) } + \text{ I am } \phi $$
... ... hmmm ... don't seem to be getting anywhere ... is $$\overline{ \phi_\alpha ( x_\alpha ) } + \text{ I am } \phi \ne 0$$ ... ?Peter

 

[steenis]

What is $\phi_\alpha$ ? As I told you in post #18, we cannot go further without knowing $\phi_\alpha$

In the posts #14, #18, #20, #23, #28, and #32, I insisted that you define $\phi_\alpha$. As long as you refuse to do so, I cannot help you, and I will stop here.
Furthermore, I told you several times what $\phi_\alpha$ is, clearly you refuse to accept that.
So come back if you are willing to define $\phi_\alpha$.

 

[Math Amateur]

What is $\phi_\alpha$ ? As I told you in post #18, we cannot go further without knowing $\phi_\alpha$

In the posts #14, #18, #20, #23, #28, and #32, I insisted that you define $\phi_\alpha$. As long as you refuse to do so, I cannot help you, and I will stop here.
Furthermore, I told you several times what $\phi_\alpha$ is, clearly you refuse to accept that.
So come back if you are willing to define $\phi_\alpha$.

Sorry steenis ... I see that you did in fact suggest a definition for $$\phi_\alpha$$ ... my apologies for forgetting this ...

You suggested defining $$\phi_g = g$$ where $$g \in \text{ Hom } ( M, N )$$ ...

so that we can construct a unique R-map ...

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

$$\phi \circ i_g = \phi_g = g $$

where $$i_g \ : \ M_g \to \bigoplus_{ \text{ Hom } ( M, N ) } M_g$$ are the canonical injections ...The constructed R-map $\phi$ is then defined on $\bigoplus_{\text{Hom }(M, N)} M_g$ in this way:

$\phi((x_g)_{\text{Hom }(M, N)}) = \Sigma_{\text{Hom }(M, N)} \text{ } g(x_g)$ for $((x_g)_{\text{Hom }(M, N)}) \in \bigoplus_{\text{Hom }(M, N)} M_g$

but ... (my apologies) ... I have an issue ... ... the family of mappings $$\{ \phi_g \ : \ M_g \to N \}$$ are already R-linear maps or R-module homomorphisms (see Proposition 2,1.5 ) ... so how does putting $$\phi_g = s$$ where $$g \in \text{ Hom } ( M, N )$$ ... that is declaring $$\phi_g$$ a member of $$\text{ Hom } ( M, N )$$ ... that is saying that $$\phi_g$$ is a homomorphism ... how does that advance our knowledge of the specific nature of the $$\phi_g$$ ... they were already homomorphisms by Proposition 2,1,5 ...Can you clarify ...

Maybe I am missing something regarding the index set being $$\text{ Hom } ( M, N )$$ ... ... Peter

 

[steenis]

Define the unknown $x$ to be the known $3$ then: $x = 3$

You define the unknown $\phi_f$ to be the known $f$ then : $\phi_f = f$, is perfectly allright

If you define $\phi_f = f$ for $f \in \text{Hom }(M, N)$ then your set $\{\phi_f = f | f \in \text{Hom }(M, N)\}$ becomes $\text{Hom }(M, N)$:

$\{\phi_f = f | f \in \text{Hom }(M, N)\} = \text{Hom }(M, N)$

Don't use s as an index-variable, we talking about R-maps so use f, g, h, or k or so.

 

[Math Amateur]

Define the unknown $x$ to be the known $3$ then: $x = 3$

You define the unknown $\phi_f$ to be the known $f$ then : $\phi_f = f$, is perfectly allright

If you define $\phi_f = f$ for $f \in \text{Hom }(M, N)$ then your set $\{\phi_f = f | f \in \text{Hom }(M, N)\}$ becomes $\text{Hom }(M, N)$:

$\{\phi_f = f | f \in \text{Hom }(M, N)\} = \text{Hom }(M, N)$

Don't use s as an index-variable, we talking about R-maps so use f, g, h, or k or so.

Changed s to g ... ... see post #48 ...

Do you have any comment about my choices for h, f and N' in post #46 ... given that now we would have $$h = \phi_g = g$$ ... ...

Peter