MHB Exact Sequences - Dummit and Foote Ch 10 - Proposition 29

[Math Amateur]

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

I am studying Proposition 29 (D&F, page 388)

I need some help in order to fully understand the proof of the last statement of Proposition 29.

Proposition 29 and its proof (Ch 10, D&F page 388) reads as follows:

View attachment 2494

As can be seen above, in the proof, D&F define $$ \pi_1 \pi_2, f, \pi_1 \circ f text{ and } \pi_2 \circ f $$ and then state the following:

"This defines a map from $$ Hom_R (D, L \oplus N) $$ to $$ Hom_R ( D, L ) \oplus Hom_R ( D, N )$$ which is easily seen to be a homomorphism.

Can someone please help me to see explicitly and formally how the definitions of MATH] \pi_1 \pi_2, f, \pi_1 \circ f text{ and } \pi_2 \circ f $$ lead to such a map and also help me to determine an explicit expression/formala for this homomorphism?

Hope someone can help.

Peter

 

[Math Amateur]

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

I am studying Proposition 29 (D&F, page 388)

I need some help in order to fully understand the proof of the last statement of Proposition 29.

Proposition 29 and its proof (Ch 10, D&F page 388) reads as follows:

View attachment 2494

As can be seen above, in the proof, D&F define $$ \pi_1 \pi_2, f, \pi_1 \circ f text{ and } \pi_2 \circ f $$ and then state the following:

"This defines a map from $$ Hom_R (D, L \oplus N) $$ to $$ Hom_R ( D, L ) \oplus Hom_R ( D, N )$$ which is easily seen to be a homomorphism.

Can someone please help me to see explicitly and formally how the definitions of MATH] \pi_1 \pi_2, f, \pi_1 \circ f text{ and } \pi_2 \circ f $$ lead to such a map and also help me to determine an explicit expression/formala for this homomorphism?

Hope someone can help.

Peter

I have reflected on my post above and believe I have a (tentative) idea of the map referred to be D&F:

I believe one should proceed as follows:

Let $$ \pi_1 : \ L \oplus N \to L $$ where $$ \pi_1 ( ( l_1, n_1)) = l_1 $$

and let $$ \pi_2 : \ L \oplus N \to N $$ where $$ \pi_2 ( ( l_1, n_1)) = n_1 $$

Now let $$ f : \ D \to L \oplus N $$ where $$ F(d) = ((l_1, n_1)) $$

Thus we have the following:

$$ \pi_1 \circ f : \ D \to L \text{ where } \pi_1( f (d) ) = \pi_1 ( ( l_1 , n_1) ) = l_1 $$

and

$$ \pi_2 \circ f : \ D \to L \text{ where } \pi_2( f (d) ) = \pi_2 ( ( l_1 , n_1) ) = n_1 $$

Now we can define what I believe to be the required homomorphism, as follows:

Let $$ F : \ Hom_R (D, L \oplus N) \to Hom_R ( D, L ) \oplus Hom_R ( D, N ) \text{ where } F(f) = ( \pi_1 \circ f , \pi_2 \circ f ) $$

Can someone please confirm that F is the required map?

Still have to show that $$F$$ is a homomorphism but I am slightly perplexed as to the details of this task.

To show $$F$$ is a module homomorphism, we must show that

$$ F(f + g) = F(f) + F(g) $$ where $$ f, g \in Hom_R (D, L \oplus N ) $$

and

$$F (rf) = rF(f)$$ for $$ r \in R $$

BUT ... how to do this?

Can someone please help?

Peter

 

[Deveno]

In categorical language, we are saying we have a FUNCTOR:

$H: R-\mathbf{Mod} \to R-\mathbf{Mod}$

given on objects by:

$H(M) = \text{Hom}_R(D,M)$

and on arrows (maps), by:

for $\psi:M \to N, H(\psi) = \psi'$, where for any $f \in \text{Hom}_R(D,M)$ we have $\psi'(f) = \psi \circ f$.

This is often called "the pullback of $\psi$ by $f$" (allowing us to "modify" $\psi$ so its domain is now "pulled back" to $D$), and the functor $H$ is often called $\text{Hom}_R(D,-)$ (an example of a "hom-functor").

If you are going to be reading more on this subject in other texts, it might be wise to assimilate some of this language now. I note in passing that this type of construction lies at the heart of the "chain rule" that proves to be such a stumbling block for many first-year calculus students. You may find it enlightening to consider what changes if we have a common "target":

$M \to \text{Hom}_R(M,D)$

**************

In any case, as with most "hom-functors", showing we have the requisite morphism properties is so "work-a-day" that the proofs are often entirely omitted. To start with, we assume an arbitrary $d \in D$. Then:

$F(f+g)(d) = [(\pi_1\circ(f+g))\times(\pi_2\circ(f+g))](d) = (\pi_1((f+g)(d)),\pi_2((f+g)(d)))$

$= (\pi_1(f(d)+g(d)),\pi_2(f(d)+g(d)) = (\pi_1(f(d)),\pi_2(f(d))) + (\pi_1(g(d)),\pi_2(g(d)))$

$= [(\pi_1 \circ f) \times (\pi_2 \circ f)](d) + [(\pi_1 \circ g) \times (p_2 \circ g)](d) = F(f)(d) + F(g)(d) = (F(f) + F(g))(d)$.

And, for any $r \in R$:

$F(rf)(d) = [(\pi_1 \circ (rf)) \times (pi_2 \circ (rf))](d) = (\pi_1((rf)(d)), \pi_2((rf)(d)))$

$= (pi_1(r(f(d))),\pi_2(r(f(d)))) = (r\pi_1(f(d)),r\pi_2(f(d))) = r(\pi_1(f(d)),\pi_2(f(d)))$

$= r[(\pi_1 \circ f) \times (\pi_2 \circ f)](d) = r(F(f(d)) = (rF)(f)(d)$.

Note that your $F$ is really just $\pi_1' \times \pi_2'$, where $\pi_1,\pi_2$ are the canonical projection homomorphisms, so in shorter language we are asserting:

$\pi_1' \times \pi_2' = (\pi_1 \times \pi_2)'$

****************

Imagine diagrams between the (single) modules lying "flat" on a surface, with $D$ lying in space above them. Each diagram is then augmented with an arrow from $D$ down to the modules lying below. The commutative triangle diagrams (from the common source $D$) we obtain from each arrow on the flat surface can be "stitched together" (composed) by identifying "matching edges".

In particular, we can make the diagram corresponding to the two projection maps:

$L \leftarrow L \oplus N \rightarrow N$

and "pulling this back to $D$", gives us two "commutative triangles" which serve the same function in the hom-sets as the projection maps do on the modules. By the universal property of the product construction:

$\text{Hom}_R(D,L \oplus N) \cong \text{Hom}_R(D,L) \oplus \text{Hom}_R(D,N)$

 

[Math Amateur]

In categorical language, we are saying we have a FUNCTOR:

$H: R-\mathbf{Mod} \to R-\mathbf{Mod}$

given on objects by:

$H(M) = \text{Hom}_R(D,M)$

and on arrows (maps), by:

for $\psi:M \to N, H(\psi) = \psi'$, where for any $f \in \text{Hom}_R(D,M)$ we have $\psi'(f) = \psi \circ f$.

This is often called "the pullback of $\psi$ by $f$" (allowing us to "modify" $\psi$ so its domain is now "pulled back" to $D$), and the functor $H$ is often called $\text{Hom}_R(D,-)$ (an example of a "hom-functor").

If you are going to be reading more on this subject in other texts, it might be wise to assimilate some of this language now. I note in passing that this type of construction lies at the heart of the "chain rule" that proves to be such a stumbling block for many first-year calculus students. You may find it enlightening to consider what changes if we have a common "target":

$M \to \text{Hom}_R(M,D)$

**************

In any case, as with most "hom-functors", showing we have the requisite morphism properties is so "work-a-day" that the proofs are often entirely omitted. To start with, we assume an arbitrary $d \in D$. Then:

$F(f+g)(d) = [(\pi_1\circ(f+g))\times(\pi_2\circ(f+g))](d) = (\pi_1((f+g)(d)),\pi_2((f+g)(d)))$

$= (\pi_1(f(d)+g(d)),\pi_2(f(d)+g(d)) = (\pi_1(f(d)),\pi_2(f(d))) + (\pi_1(g(d)),\pi_2(g(d)))$

$= [(\pi_1 \circ f) \times (\pi_2 \circ f)](d) + [(\pi_1 \circ g) \times (p_2 \circ g)](d) = F(f)(d) + F(g)(d) = (F(f) + F(g))(d)$.

And, for any $r \in R$:

$F(rf)(d) = [(\pi_1 \circ (rf)) \times (pi_2 \circ (rf))](d) = (\pi_1((rf)(d)), \pi_2((rf)(d)))$

$= (pi_1(r(f(d))),\pi_2(r(f(d)))) = (r\pi_1(f(d)),r\pi_2(f(d))) = r(\pi_1(f(d)),\pi_2(f(d)))$

$= r[(\pi_1 \circ f) \times (\pi_2 \circ f)](d) = r(F(f(d)) = (rF)(f)(d)$.

Note that your $F$ is really just $\pi_1' \times \pi_2'$, where $\pi_1,\pi_2$ are the canonical projection homomorphisms, so in shorter language we are asserting:

$\pi_1' \times \pi_2' = (\pi_1 \times \pi_2)'$

****************

Imagine diagrams between the (single) modules lying "flat" on a surface, with $D$ lying in space above them. Each diagram is then augmented with an arrow from $D$ down to the modules lying below. The commutative triangle diagrams (from the common source $D$) we obtain from each arrow on the flat surface can be "stitched together" (composed) by identifying "matching edges".

In particular, we can make the diagram corresponding to the two projection maps:

$L \leftarrow L \oplus N \rightarrow N$

and "pulling this back to $D$", gives us two "commutative triangles" which serve the same function in the hom-sets as the projection maps do on the modules. By the universal property of the product construction:

$\text{Hom}_R(D,L \oplus N) \cong \text{Hom}_R(D,L) \oplus \text{Hom}_R(D,N)$

Thanks so much for the help Deveno ... Just working through your post now ...

Peter

 

[Math Amateur]

In categorical language, we are saying we have a FUNCTOR:

$H: R-\mathbf{Mod} \to R-\mathbf{Mod}$

given on objects by:

$H(M) = \text{Hom}_R(D,M)$

and on arrows (maps), by:

for $\psi:M \to N, H(\psi) = \psi'$, where for any $f \in \text{Hom}_R(D,M)$ we have $\psi'(f) = \psi \circ f$.

This is often called "the pullback of $\psi$ by $f$" (allowing us to "modify" $\psi$ so its domain is now "pulled back" to $D$), and the functor $H$ is often called $\text{Hom}_R(D,-)$ (an example of a "hom-functor").

If you are going to be reading more on this subject in other texts, it might be wise to assimilate some of this language now. I note in passing that this type of construction lies at the heart of the "chain rule" that proves to be such a stumbling block for many first-year calculus students. You may find it enlightening to consider what changes if we have a common "target":

$M \to \text{Hom}_R(M,D)$

**************

In any case, as with most "hom-functors", showing we have the requisite morphism properties is so "work-a-day" that the proofs are often entirely omitted. To start with, we assume an arbitrary $d \in D$. Then:

$F(f+g)(d) = [(\pi_1\circ(f+g))\times(\pi_2\circ(f+g))](d) = (\pi_1((f+g)(d)),\pi_2((f+g)(d)))$

$= (\pi_1(f(d)+g(d)),\pi_2(f(d)+g(d)) = (\pi_1(f(d)),\pi_2(f(d))) + (\pi_1(g(d)),\pi_2(g(d)))$

$= [(\pi_1 \circ f) \times (\pi_2 \circ f)](d) + [(\pi_1 \circ g) \times (p_2 \circ g)](d) = F(f)(d) + F(g)(d) = (F(f) + F(g))(d)$.

And, for any $r \in R$:

$F(rf)(d) = [(\pi_1 \circ (rf)) \times (pi_2 \circ (rf))](d) = (\pi_1((rf)(d)), \pi_2((rf)(d)))$

$= (pi_1(r(f(d))),\pi_2(r(f(d)))) = (r\pi_1(f(d)),r\pi_2(f(d))) = r(\pi_1(f(d)),\pi_2(f(d)))$

$= r[(\pi_1 \circ f) \times (\pi_2 \circ f)](d) = r(F(f(d)) = (rF)(f)(d)$.

Note that your $F$ is really just $\pi_1' \times \pi_2'$, where $\pi_1,\pi_2$ are the canonical projection homomorphisms, so in shorter language we are asserting:

$\pi_1' \times \pi_2' = (\pi_1 \times \pi_2)'$

****************

Imagine diagrams between the (single) modules lying "flat" on a surface, with $D$ lying in space above them. Each diagram is then augmented with an arrow from $D$ down to the modules lying below. The commutative triangle diagrams (from the common source $D$) we obtain from each arrow on the flat surface can be "stitched together" (composed) by identifying "matching edges".

In particular, we can make the diagram corresponding to the two projection maps:

$L \leftarrow L \oplus N \rightarrow N$

and "pulling this back to $D$", gives us two "commutative triangles" which serve the same function in the hom-sets as the projection maps do on the modules. By the universal property of the product construction:

$\text{Hom}_R(D,L \oplus N) \cong \text{Hom}_R(D,L) \oplus \text{Hom}_R(D,N)$

Thanks again for the help Deveno ...

Can you demonstrate explicitly why the following step in your reasoning is justified:

$= (\pi_1(f(d)+g(d)),\pi_2(f(d)+g(d)) = (\pi_1(f(d)),\pi_2(f(d))) + (\pi_1(g(d)),\pi_2(g(d)))$

Would appreciate help regarding the reason why this step is valid.

Peter

 

[Math Amateur]

Thanks again for the help Deveno ...

Can you demonstrate explicitly why the following step in your reasoning is justified:

$= (\pi_1(f(d)+g(d)),\pi_2(f(d)+g(d)) =

Would appreciate help regarding the reason why this step is valid.

Peter

Just been reflecting on my own question ... and believe I have seen the reason (which I have to say appears to be quite straightforward ...)

We have

$$= (\pi_1(f(d)+g(d)),\pi_2(f(d)+g(d)) = (\pi_1(f(d)) + \pi_1(g(d), \pi_2(f(d)) + \pi_2(g(d)))$$

$$ = (\pi_1(f(d)),\pi_2(f(d))) + (\pi_1(g(d)),\pi_2(g(d))) $$

since $$ (x_1 + x_2, y_1 + y_2) = (x_1, y_1) + (x_2, y_2) $$ by the definition of component-wise addition.

Can you confirm this is OK?

It is pretty simple and looks OK ... but just to be sure ...

Peter

 

[Deveno]

That is correct.