MHB Free Abelian Groups .... Aluffi Proposition 5.6

[Math Amateur]

I am reading Paolo Aluffi's book: Algebra: Chapter 0 ... ...

I am currently focussed on Section 5.4 Free Abelian Groups ... ...

I need help with an aspect of Aluffi's preamble to introduce Proposition 5.6 ...

Proposition 5.6 and its preamble reads as follows:
View attachment 5582In the above text from Aluffi's book we find the following:

" ... ... For $$H = \mathbb{Z}$$ there is a natural function $$j \ : \ A \longrightarrow \mathbb{Z}^{ \oplus A }$$ , obtained by sending $$a \in A$$ to the function $$j_a \ : \ A \longrightarrow \mathbb{Z}$$ ... ... "

My problem is in (precisely and rigorously) understanding the claim that $$j$$ sends $$a \in A$$ to the function $$j_a \ : \ A \longrightarrow \mathbb{Z}$$ ... ...

The function $$j_a$$ is actually a set of ordered pairs no two of which have the same member ... BUT ... $$j$$ does not (exactly anyway) seem to send $$a \in A$$ to this function ...If we pretend for a moment that $$A$$ is a countable ordered set ... then we can say that what $$j$$ seems to do is send $$a \in A$$ to the image-set of $$j_a$$, namely

$$( \ ... \ ... \ ,0,0, 0, \ ... \ ... \ 0,1,0 \ ... \ ... \ ... \ \ ,0,0, 0, \ ... \ ... ) $$

where the 1 is in the a'th position ...

So then $$j$$ seems to send $$a \in A$$ to the image-set of $$j_a$$ and not to $$j_a$$ itself ... ... (... ... not sure how to put this argument for the case where the set A is uncountable and not ordered ... ... )

Given my analysis ... how do we justify or make sense of Aluffi's claim that " there is a natural function $$j \ : \ A \longrightarrow \mathbb{Z}^{ \oplus A }$$ , obtained by sending $$a \in A$$ to the function $$j_a \ : \ A \longrightarrow \mathbb{Z}$$ "

Hope someone can critique my analysis and clarify the issue to which I refer ...

Help will be much appreciated ...

Peter*** EDIT ***

Just another concern over possibly missing something in fully understanding Aluffi's text above ... he introduces the general case with a general abelian group $$H$$ ... ... and then defines $$H^{ \oplus A}$$ ... ... but never uses $$H$$ ... he just puts it equal to $$\mathbb{Z}$$ ... if you are just going to put $$H = \mathbb{Z}$$ ... ... why bother with $$H$$ ... why introduce it ... just start with $$\mathbb{Z}$$ ... ... does anyone have an idea what Aluffi is doing ... ... ? ... ... he does a similar thing when explaining free modules ... ... am I missing something ... ... ? ... ...
======================================================

To give MHB members reading this post a sense of the approach and notation of Aluffi to free Abelian groups I am here providing Aluffi's introduction to free Abelian groups up to and including Proposition 5.6 ... as follows:View attachment 5583
View attachment 5584
View attachment 5585

 

[Deveno]

Let's look at the simplest non-trivial example, when $n = 2$, and $A = \{1,2\}$.

In the *product* $\Bbb Z \times \Bbb Z$, we see an element as a pair: $(a,b): a,b \in \Bbb Z$. These pairs should be thought as a "single things", much like a point in the plane is a "single thing" quite different than two real numbers "pushed together".

The process of "splitting" the pair $(a,b)$ into its two *components", $a$ and $b$, is a "reduction process" we get by ignoring part of the pair.

The *co-product* is different, at least in spirit-instead of a "big thing" that we can "split factors off of", it is the factors "minimally embedded" in a larger thing that acts as a whole. The process of creating a co-product should be thought of as "amalgamating" in some regard (like putting together hot fudge and ice cream to make a sundae).

In the co-product $\Bbb Z^{\oplus 2}$, or as we might write $\Bbb Z \oplus \Bbb Z$, Aluffi is trying to establish a similarity in its construction with the disjoint union of sets-recall that in the set-product, we just take the cartesian product (pairs of elements), but in the set co-product, we take pairs of disjoint (by construction) elements.

For example, in $A \times A$, there is just one "diagonal element" $(a,a)$ for any $a \in A$. In $A \amalg A$ there are two such elements: $(a,1)$ and $(a,2)$-indeed the disjoint union can be thought of as the union of the two sets:

$\{f: \{1\} \to A\} \cup \{g:\{2\} \to A\}$ (it is clear this union is disjoint).

In this case, letting $2$ stand for the set $\{1,2\}$, we can form the set $\text{Hom}_{\mathbf{Set}}(2,\Bbb Z) = \Bbb Z^2$ as all functions (which of necessity have finite support):

$f: {1,2} \to \Bbb Z$.

For example, here is one such function:

$1 \mapsto a$
$2 \mapsto b$

We add these functions "point-wise", given $f$ and $g$, we say that $f+g$ is THIS function:

$1 \mapsto f(1) + g(1)$
$2 \mapsto f(2) + g(2)$.

This agrees with the common *definition* of an ordered pair in a given set $A$ (an element of $A \times A$) as a function $f:{1,2} \to A$, we can simply identify the function $f$:

$1 \mapsto a$
$2 \mapsto b$

with the ordered pair $(a,b)$. This is fine...EXCEPT when we start generalizing to infinite sets.

For a countable well-ordered set (think: the natural numbers) $W$, we can define our functions as $W$-sequences, and write:

$f = (a_0,a_1,a_2,\dots)$ where $a_i = f(w_i)$.

But if we want to write such things as SUMS (using the abelian group operation of the image-set of the $f$'s) of such things as:

$(1,0,0,\dots)$
$(0,1,0,\dots)$
$(0,0,\dots,1,0,\dots)$

to express a constant sequence like: $(1,1,1,\dots)$, we will require an INFINITE number of summands.

On the other hand, the things we can express by FINITE sums, is going to be a smaller set of functions. The direct product of countably infinitely many copies of $\Bbb Z$ has sequences that "go on forever", but the direct sum has only sequences that eventually terminate. That is, when you GENERALIZE the direct product and direct sum to INFINITE indexing sets, you will, in general, get two different abelian groups (the direct product will CONTAIN (an isomorphic copy of) the direct sum, but not vice-versa).

Now the reason Aluffi focuses in on just $\Bbb Z$, rather than an arbitrary abelian group $H$, is because $\Bbb Z$ plays a central role in abelian groups (look up the structure theorem), and also:

A free abelian group is the same thing as a free $\Bbb Z$-module (compare the construction mimicking Cooperstein of "the $\Bbb Z$-module based on the set $\{1,2,\dots,n\}$" and its UMP with the UMP presented here).

I don't know if I'm making this *clear* enough but YOU are the one identifying $j_a$ with its image-set (it's images DO NOT lie in $\Bbb Z^{\oplus n}$, they lie in $\Bbb Z$). If $A$ is NOT a finite set, then the functions with "finite support" on $A$ become an important subset.

 

[Math Amateur]

Let's look at the simplest non-trivial example, when $n = 2$, and $A = \{1,2\}$.

In the *product* $\Bbb Z \times \Bbb Z$, we see an element as a pair: $(a,b): a,b \in \Bbb Z$. These pairs should be thought as a "single things", much like a point in the plane is a "single thing" quite different than two real numbers "pushed together".

The process of "splitting" the pair $(a,b)$ into its two *components", $a$ and $b$, is a "reduction process" we get by ignoring part of the pair.

The *co-product* is different, at least in spirit-instead of a "big thing" that we can "split factors off of", it is the factors "minimally embedded" in a larger thing that acts as a whole. The process of creating a co-product should be thought of as "amalgamating" in some regard (like putting together hot fudge and ice cream to make a sundae).

In the co-product $\Bbb Z^{\oplus 2}$, or as we might write $\Bbb Z \oplus \Bbb Z$, Aluffi is trying to establish a similarity in its construction with the disjoint union of sets-recall that in the set-product, we just take the cartesian product (pairs of elements), but in the set co-product, we take pairs of disjoint (by construction) elements.

For example, in $A \times A$, there is just one "diagonal element" $(a,a)$ for any $a \in A$. In $A \amalg A$ there are two such elements: $(a,1)$ and $(a,2)$-indeed the disjoint union can be thought of as the union of the two sets:

$\{f: \{1\} \to A\} \cup \{g:\{2\} \to A\}$ (it is clear this union is disjoint).

In this case, letting $2$ stand for the set $\{1,2\}$, we can form the set $\text{Hom}_{\mathbf{Set}}(2,\Bbb Z) = \Bbb Z^2$ as all functions (which of necessity have finite support):

$f: {1,2} \to \Bbb Z$.

For example, here is one such function:

$1 \mapsto a$
$2 \mapsto b$

We add these functions "point-wise", given $f$ and $g$, we say that $f+g$ is THIS function:

$1 \mapsto f(1) + g(1)$
$2 \mapsto f(2) + g(2)$.

This agrees with the common *definition* of an ordered pair in a given set $A$ (an element of $A \times A$) as a function $f:{1,2} \to A$, we can simply identify the function $f$:

$1 \mapsto a$
$2 \mapsto b$

with the ordered pair $(a,b)$. This is fine...EXCEPT when we start generalizing to infinite sets.

For a countable well-ordered set (think: the natural numbers) $W$, we can define our functions as $W$-sequences, and write:

$f = (a_0,a_1,a_2,\dots)$ where $a_i = f(w_i)$.

But if we want to write such things as SUMS (using the abelian group operation of the image-set of the $f$'s) of such things as:

$(1,0,0,\dots)$
$(0,1,0,\dots)$
$(0,0,\dots,1,0,\dots)$

to express a constant sequence like: $(1,1,1,\dots)$, we will require an INFINITE number of summands.

On the other hand, the things we can express by FINITE sums, is going to be a smaller set of functions. The direct product of countably infinitely many copies of $\Bbb Z$ has sequences that "go on forever", but the direct sum has only sequences that eventually terminate. That is, when you GENERALIZE the direct product and direct sum to INFINITE indexing sets, you will, in general, get two different abelian groups (the direct product will CONTAIN (an isomorphic copy of) the direct sum, but not vice-versa).

Now the reason Aluffi focuses in on just $\Bbb Z$, rather than an arbitrary abelian group $H$, is because $\Bbb Z$ plays a central role in abelian groups (look up the structure theorem), and also:

A free abelian group is the same thing as a free $\Bbb Z$-module (compare the construction mimicking Cooperstein of "the $\Bbb Z$-module based on the set $\{1,2,\dots,n\}$" and its UMP with the UMP presented here).

I don't know if I'm making this *clear* enough but YOU are the one identifying $j_a$ with its image-set (it's images DO NOT lie in $\Bbb Z^{\oplus n}$, they lie in $\Bbb Z$). If $A$ is NOT a finite set, then the functions with "finite support" on $A$ become an important subset.

Thanks Deveno ... still reflecting on everything you have said ... great that you explained via simple examples ... I found that that was extremely helpful and illuminating ...

I hope that ... without trying your patience ... I can ask a further question ...

Despite the clarification at the end of your post I am still somewhat puzzled about how j works exactly ... indeed I wrote previously ... sorry to be a bit slow and pedestrian here ... ... :( ... ... " ... ... My problem is in (precisely and rigorously) understanding the claim that $$j$$ sends $$a \in A$$ to the function $$j_a \ : \ A \longrightarrow \mathbb{Z}$$ ... ...

The function $$j_a$$ is actually a set of ordered pairs no two of which have the same member ... BUT ... $$j$$ does not (exactly anyway) seem to send $$a \in A$$ to this function ..."
So ... I am still perplexed about the nature and working of the function j ... ... basically how does j work in a simple example ... how does it send $$a \in A$$ to a function ... and how does the mechanics of this work in the scheme of things ... ... ?

Are you able to help ... ?

Peter

 

[Deveno]

OK, we have an abelian group (in this case, the integers).

Next, we have a function space:

$\text{Hom}_{\mathbf{Set}}(\{1,2\},\Bbb Z\}$

a typical element of which looks like this:

$f:\{1,2\} \to \Bbb Z$, where $f(1) = k$, and $f(2) = m$ ($k$ and $m$ are integers).

Given two functions $f,g$ we DEFINE a sum using the addition of integers:

$(f+g)(1) = f(1) + g(1)$
$(f+g)(2) = f(2) + g(2)$.

So if $g(1) = k'$ and $g(2) = m'$, then $f+g$ does this:

$1 \mapsto k+k'$
$2 \mapsto m+m'$.

This set of functions with this addition forms an abelian group:

The operation is associative:

$[(f+g) + h](1) = (f+g)(1) + h(1) = (f(1) + g(1)) + h(1) = f(1) + (g(1) + h(1)) = f(1) + (g+h)(1) = [f + (g+h)](1)$

and similarly $(f+g) + h](2) = [f+(g+h)](2)$ so that $(f+g) + h = f + (g+h)$.

The function $z:1 \mapsto 0, 2 \mapsto 0$ functions as an identity.

For any $f$, the function $-f$ defined by:

$-f(i) = -(f(i))$ is an additive inverse, and our addition is abelian (proven in much the same way as associativity).

Now here $A = \{1,2\}$, and the function $j$ is given by:

$j(1) = j_1$
$j(2) = j_2$, and

$j_1(1) = 1$
$j_2(2) = 0$

$j_2(1) = 0$
$j_2(2) = 1$

In other words $j_i(k) = \delta_{ik}$, the kronecker delta, which would be the identity 2x2 matrix if we wrote out all possible values of $j_i(k)$ as a two-dimensional array, where each COLUMN is indexed by $i$ and each ROW is indexed by $k$.

What we are considering here is finite $\Bbb Z$-linear combinations of the $j_i$ (remember everything we are doing lives in $\text{Hom}_{\mathbf{Set}}(\{1,2\},\Bbb Z)$).

Now, clearly there is a BIJECTION between these two sets:

$\{j_1,j_2\}$ and $\{(1,0),(0,1)\}$, and furthermore both are a $\Bbb Z$-BASIS of a free $\Bbb Z$-module of rank 2. In fact, this bijection actually induces a $\Bbb Z$-module (that is abelian group) isomorphism, given by:

$a_1j_1 + a_2j_2 \mapsto (a_1,a_2)$

But look at the difference in FORM: the left is a sum of functions, the right is just a pair.

This difference in form is inconsequential when our indexing set $A$ is finite. When $A$ is infinite, this is not so-addition only "makes sense" when we have a finite number of things to add (how do we add "infinite things", except formally-there's a problem of EVALUATION of the functions, then), and then it's not a bijection, anymore.

 

[Math Amateur]

OK, we have an abelian group (in this case, the integers).

Next, we have a function space:

$\text{Hom}_{\mathbf{Set}}(\{1,2\},\Bbb Z\}$

a typical element of which looks like this:

$f:\{1,2\} \to \Bbb Z$, where $f(1) = k$, and $f(2) = m$ ($k$ and $m$ are integers).

Given two functions $f,g$ we DEFINE a sum using the addition of integers:

$(f+g)(1) = f(1) + g(1)$
$(f+g)(2) = f(2) + g(2)$.

So if $g(1) = k'$ and $g(2) = m'$, then $f+g$ does this:

$1 \mapsto k+k'$
$2 \mapsto m+m'$.

This set of functions with this addition forms an abelian group:

The operation is associative:

$[(f+g) + h](1) = (f+g)(1) + h(1) = (f(1) + g(1)) + h(1) = f(1) + (g(1) + h(1)) = f(1) + (g+h)(1) = [f + (g+h)](1)$

and similarly $(f+g) + h](2) = [f+(g+h)](2)$ so that $(f+g) + h = f + (g+h)$.

The function $z:1 \mapsto 0, 2 \mapsto 0$ functions as an identity.

For any $f$, the function $-f$ defined by:

$-f(i) = -(f(i))$ is an additive inverse, and our addition is abelian (proven in much the same way as associativity).

Now here $A = \{1,2\}$, and the function $j$ is given by:

$j(1) = j_1$
$j(2) = j_2$, and

$j_1(1) = 1$
$j_2(2) = 0$

$j_2(1) = 0$
$j_2(2) = 1$

In other words $j_i(k) = \delta_{ik}$, the kronecker delta, which would be the identity 2x2 matrix if we wrote out all possible values of $j_i(k)$ as a two-dimensional array, where each COLUMN is indexed by $i$ and each ROW is indexed by $k$.

What we are considering here is finite $\Bbb Z$-linear combinations of the $j_i$ (remember everything we are doing lives in $\text{Hom}_{\mathbf{Set}}(\{1,2\},\Bbb Z)$).

Now, clearly there is a BIJECTION between these two sets:

$\{j_1,j_2\}$ and $\{(1,0),(0,1)\}$, and furthermore both are a $\Bbb Z$-BASIS of a free $\Bbb Z$-module of rank 2. In fact, this bijection actually induces a $\Bbb Z$-module (that is abelian group) isomorphism, given by:

$a_1j_1 + a_2j_2 \mapsto (a_1,a_2)$

But look at the difference in FORM: the left is a sum of functions, the right is just a pair.

This difference in form is inconsequential when our indexing set $A$ is finite. When $A$ is infinite, this is not so-addition only "makes sense" when we have a finite number of things to add (how do we add "infinite things", except formally-there's a problem of EVALUATION of the functions, then), and then it's not a bijection, anymore.

Well ... that was extremely helpful ... worked through that a few times ...

Thanks so much for clarifying that issue ... none of the texts I inspected helped ...

Your posts make some difficulties just disappear ... the examples are really helpful in getting a real understanding of difficult notions ...

Thanks again,

Peter