MHB Exact Sequences - D&F Section 10.5, Proposition 27

[Math Amateur]

I am reading Dummit and Foote Section 10.5 Exact Sequences - Projective, Injective and Flat Modules.

I need some help in understanding D&F's proof of Proposition 27, Section 10.5, page 386 (see attachment).

Proposition 27 reads as follows: (see attachment)

------------------------------------------------------------------------------

Let D, L and M be R-modules and let

$$ \psi \ : \ L \to M $$

be an R-module homomorphism.

Then the map

$$ {\psi}' \ : \ Hom_R (D, L) \to Hom_R (D, M) $$

where $$ f \to f' = \psi \circ f $$

is a homomorphism of abelian groups. ... ...

... ... ... "

---------------------------------------------------------------------------

D&F start their proof of the proposition as follows:

"The fact that $$ {\psi}' $$ is a homomorphism is immediate. ... ... "Can someone please help me formulate an explicit, formal and rigorous proof of this proposition (however short and obvious it is!).
My tentative thoughts are as follows:

$$ {\psi}' \ : \ f \to f' $$ is a homomorphism

$$ \Longrightarrow {\psi}' (f_1 \circ f_2) = {\psi}' (f_1) \cdot {\psi}' (f_2) $$

(where $$ \circ $$ and $$ \cdot $$ are both the operation of composition of functions)

Now $$ {\psi}' (f_1 \circ f_2) = \psi \circ (f_1 \circ f_2) $$

Then I want to write:

$$ \psi \circ (f_1 \circ f_2) = ( \psi \circ f_1 ) \circ ( \psi \circ f_2 ) $$

but what would be the justification for this step?

... ... can some please help me in the task of formulating a rigorous proof?

Peter

 

[Math Amateur]

I am reading Dummit and Foote Section 10.5 Exact Sequences - Projective, Injective and Flat Modules.

I need some help in understanding D&F's proof of Proposition 27, Section 10.5, page 386 (see attachment).

Proposition 27 reads as follows: (see attachment)

------------------------------------------------------------------------------

Let D, L and M be R-modules and let

$$ \psi \ : \ L \to M $$

be an R-module homomorphism.

Then the map

$$ {\psi}' \ : \ Hom_R (D, L) \to Hom_R (D, M) $$

where $$ f \to f' = \psi \circ f $$

is a homomorphism of abelian groups. ... ...

... ... ... "

---------------------------------------------------------------------------

D&F start their proof of the proposition as follows:

"The fact that $$ {\psi}' $$ is a homomorphism is immediate. ... ... "Can someone please help me formulate an explicit, formal and rigorous proof of this proposition (however short and obvious it is!).
My tentative thoughts are as follows:

$$ {\psi}' \ : \ f \to f' $$ is a homomorphism

$$ \Longrightarrow {\psi}' (f_1 \circ f_2) = {\psi}' (f_1) \cdot {\psi}' (f_2) $$

(where $$ \circ $$ and $$ \cdot $$ are both the operation of composition of functions)

Now $$ {\psi}' (f_1 \circ f_2) = \psi \circ (f_1 \circ f_2) $$

Then I want to write:

$$ \psi \circ (f_1 \circ f_2) = ( \psi \circ f_1 ) \circ ( \psi \circ f_2 ) $$

but what would be the justification for this step?

... ... can some please help me in the task of formulating a rigorous proof?

Peter

I think I have clarified my thinking ... hopefully someone will confirm this is the case ...

I began wondering about the operation between elements in the abelian group $$ Hom_R (D,L) $$ ... it is necessary to have this clear when considering the map

$$ {\psi}' \ : \ Hom_R (D, L) \to Hom_R (D, M) $$

where $$ f \to f' = \psi \circ f $$

as a homomorphism.

So, basically when we consider the defining equation of a group homomorphism in this case - visually:

$$ {\psi}' (f_1 \circ f_2) = {\psi}' (f_1) \cdot {\psi}' (f_2) $$

we need to examine the binary operation in the abelian group $$ Hom_R (D,L) $$ and following D&F Proposition 2, Section 10.2 we have, given two elements $$ f_1 \text{ and } f_2 \in Hom_R (D,L) $$:

$$ (f_1 +_D f_2) (d) = f_1(d) +_L f_2 (d) \text{ for all } d \in D $$

So when considering $$ {\psi}' $$ we should write:

$$ {\psi}' $$ is a homomorphism

$$ \Longrightarrow {\psi}' (f_1 +_D f_2) = {\psi}' (f_1) +_L {\psi}' (f_2) $$

Now we have

$$ {\psi}' (f_1 +_D f_2) = \psi \circ ( f_1 +_D f_2) $$

Can someone confirm that this revision in my thinking is correct?

I now wish to proceed to say that:

$$ \psi \circ ( f_1 +_D f_2) = (\psi \circ f_1) +_M (\psi \circ f_1) = {\psi}' (f_1) +_M {\psi}' (f_2) $$

But how do I justify this step - maybe I have to examine functional values? Can someone please help?

Peter

 

[Deveno]

The operation in $\text{Hom}_R(D,L)$ is "pointwise addition" that is:

$(f_1 + f_2)(d) = f_1(d) +_L f_2(d)$ for $f_1,f_2 \in \text{Hom}_R(D,L), d \in D$.

(the addition in $\text{Hom}_R(D,L)$ takes place in that group, NOT in $D$).

The only time "addition in $D$" comes into the picture is when we verify that $f_1 + f_2$ is indeed a member of $\text{Hom}_R(D,L)$ (that is, show it is $R$-linear).

So you must verify that:

$\psi'(f_1 + f_2) = \psi'(f_1) + \psi'(f_2)$.

The addition on the left takes place in $\text{Hom}_R(D,L)$, the addition on the right takes place in $\text{Hom}_R(D,M)$.

First, we might want to show explicitly, that $\psi'(f) = \psi \circ f$ is actually IN $\text{Hom}_R(D,M)$.

To do this, pick any two arbitrary elements $d_1,d_2 \in D$ with $r$ any element of $R$.

Then:

$\psi'(f)(d_1 +_D d_2) = \psi(f(d_1 +_D d_2)) = \psi(f(d_1) +_L f(d_2))$

$ = \psi(f(d_1)) +_M \psi(f(d_2)) = \psi'(f)(d_1) +_M \psi'(f)(d_2)$

so $\psi'(f)$ is certainly an additive map ( that is, is in $\text{Hom}_{\ \Bbb Z}(D,M)$).

and furthermore:

$\psi'(f)(r\cdot d_1) = \psi(f(r\cdot d_1)) = \psi(r\cdot f(d_1))$

$= r\cdot \psi(f(d_1)) = r \cdot (\psi'(f)(d_1))$, so $\psi'(f)$ is $R$-linear.

Now show that for any $d \in D$ that:

$\psi'(f_1 + f_2)(d) = \psi'(f_1)(d) + \psi'(f_2)(d)$

(this equality takes place in $M$), which will show that:

$\psi'(f_1 + f_2)$ and $\psi'(f_1) + \psi'(f_2)$ are the same element of $\text{Hom}_R(D,M)$.

(This is quite straight-forward, everything goes through as you would expect).

 

[Math Amateur]

The operation in $\text{Hom}_R(D,L)$ is "pointwise addition" that is:

$(f_1 + f_2)(d) = f_1(d) +_L f_2(d)$ for $f_1,f_2 \in \text{Hom}_R(D,L), d \in D$.

(the addition in $\text{Hom}_R(D,L)$ takes place in that group, NOT in $D$).

The only time "addition in $D$" comes into the picture is when we verify that $f_1 + f_2$ is indeed a member of $\text{Hom}_R(D,L)$ (that is, show it is $R$-linear).

So you must verify that:

$\psi'(f_1 + f_2) = \psi'(f_1) + \psi'(f_2)$.

The addition on the left takes place in $\text{Hom}_R(D,L)$, the addition on the right takes place in $\text{Hom}_R(D,M)$.

First, we might want to show explicitly, that $\psi'(f) = \psi \circ f$ is actually IN $\text{Hom}_R(D,M)$.

To do this, pick any two arbitrary elements $d_1,d_2 \in D$ with $r$ any element of $R$.

Then:

$\psi'(f)(d_1 +_D d_2) = \psi(f(d_1 +_D d_2)) = \psi(f(d_1) +_L f(d_2))$

$ = \psi(f(d_1)) +_M \psi(f(d_2)) = \psi'(f)(d_1) +_M \psi'(f)(d_2)$

so $\psi'(f)$ is certainly an additive map ( that is, is in $\text{Hom}_{\ \Bbb Z}(D,M)$).

and furthermore:

$\psi'(f)(r\cdot d_1) = \psi(f(r\cdot d_1)) = \psi(r\cdot f(d_1))$

$= r\cdot \psi(f(d_1)) = r \cdot (\psi'(f)(d_1))$, so $\psi'(f)$ is $R$-linear.

Now show that for any $d \in D$ that:

$\psi'(f_1 + f_2)(d) = \psi'(f_1)(d) + \psi'(f_2)(d)$

(this equality takes place in $M$), which will show that:

$\psi'(f_1 + f_2)$ and $\psi'(f_1) + \psi'(f_2)$ are the same element of $\text{Hom}_R(D,M)$.

(This is quite straight-forward, everything goes through as you would expect).

Thanks for a really clear and helpful post, Deveno

Appreciate the help,

Peter