MHB Universal Properties - Quotients - Aluffi, Chapter 1, Section 5.

[Math Amateur]

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

I am, at present, focused on Chapter 1, Section 5: Universal Properties.

I subsection 5.3 on Quotients, Aluffi writes the following:
View attachment 2621I am uncertain of the nature of the category that Aluffi is constructing. in particular, do the functions $$ \phi_1 \text{ and } \phi_2 $$ represent different equivalence relations $$ \sim_1 , \sim_2 $$? (if not what do they represent?)

It seems to me that they do represent/concern different equivalence relations ... BUT on the next page, (following on from the text displayed above) Aluffi writes:
View attachment 2623
Given that Aluffi is here talking about the pair $$ ( \pi , A/ \sim ) $$ it seems that the equivalence relation $$ \sim $$ is quite fundamental ... so maybe the functions $$ \phi_1 \text{ and } \phi_2 $$ do not represent different equivalence relations ...

Can someone please clarify the nature of the category that Aluffi is describing?

Would appreciate the help.

Peter

 

[Deveno]

First let's be clear how we are constructing the category Aluffi means:

We start with a set $A$ and an equivalence relation $\sim$ on $A$.

Now the OBJECTS of our category, are ARROWS of $\mathbf{Set}$, in fact the objects under consideration all have a common source ("domain"), the set $A$. Since the objects all have source $A$, we can think of them as a pair: $(Z,\varphi)$.

But not just "any" functions $\varphi:A \to Z$, we only want the functions that induce "well-defined" functions on the set of equivalence classes: $A/\sim$ (these functions are said to RESPECT the equivalence).

It should be clear that even if we talk about the same "target" (co-domain) $Z$, we might get several such functions, for example any CONSTANT function $A \to Z$ will respect the equivalence $\sim$.

I also want to make it clear that we are considering the equivalence relation $\sim$ as FIXED, just like the set $A$.

OK, so we now what the OBJECTS of our category are, what are the ARROWS?

An arrow from $(Z_1,\varphi_1)$ to $(Z_2,\varphi_2)$ is an arrow $\sigma: Z_1 \to Z_2$ such that:

$\varphi_2 = \sigma \circ \varphi_1$

It should be clear that this is exactly the same information encoded by the commutative diagram, so we can think of arrows in our category as these diagrams.

To verify this is a bona-fide category, we need to show 2 things:

1. We have identity arrows.
2. Arrows are composable.

It should be clear that the arrow $1_Z$ serves as an identity arrow for the pair $(Z,\varphi)$. To see that arrows are composable, we join two commutative diagrams along "the matching arrow in the middle".

What Aluffi is doing here, is easing you into the notion of a co-slice category, which in turn is a special case of a "comma category". In this case, we can think of arrows as functions $\tilde{\varphi}:A/\sim \to Z$, and our category is $(A/\sim\downarrow\mathbf{Set})$, or: "the sets under $A/\sim$".

Now what would it mean for an object $(I,f)$ to be initial in this category? It would mean there is a UNIQUE arrow $(I,f) \to (Z,\varphi)$. Remember, $(I,f)$ really means an arrow:

$f:A \to I$ such that $f(a') = f(a)$ for all $a' \sim a$.

Now if: $I = A/\sim$ and $f = \pi$, where $\pi(a) = [a]$ (the equivalence class of $a$), by the definition of ARROWS in our category, we have that an arrow from $(A/\sim,\pi) \to (Z,\varphi)$ is a function:

$g:A/\sim \to Z$ such that $g \circ \pi = \varphi$.

It should (hopefully) be clear that we HAVE to have $g = \tilde{\varphi}$ where:

$\tilde{\varphi}([a]) = \varphi(a)$, for all $a \in A$, by commutativity of the relevant diagram.

In other words, we can induce a well-defined function $\tilde{\varphi}$ for any function $\varphi:A \to Z$ that respects $\sim$ by factoring $\varphi$ through $\pi$.

If you want to "bring this down to earth", suppose that $A = \Bbb Z$, and that we require that all our maps be group homomorphisms. Define $\sim$ by:

$k \sim m \iff k - m \in n\Bbb Z$.

Then "reducing mod $n$" is universal among all group homomorphisms out of the integers that annihilate the multiples of $n$.

 

[Math Amateur]

First let's be clear how we are constructing the category Aluffi means:

We start with a set $A$ and an equivalence relation $\sim$ on $A$.

Now the OBJECTS of our category, are ARROWS of $\mathbf{Set}$, in fact the objects under consideration all have a common source ("domain"), the set $A$. Since the objects all have source $A$, we can think of them as a pair: $(Z,\varphi)$.

But not just "any" functions $\varphi:A \to Z$, we only want the functions that induce "well-defined" functions on the set of equivalence classes: $A/\sim$ (these functions are said to RESPECT the equivalence).

It should be clear that even if we talk about the same "target" (co-domain) $Z$, we might get several such functions, for example any CONSTANT function $A \to Z$ will respect the equivalence $\sim$.

I also want to make it clear that we are considering the equivalence relation $\sim$ as FIXED, just like the set $A$.

OK, so we now what the OBJECTS of our category are, what are the ARROWS?

An arrow from $(Z_1,\varphi_1)$ to $(Z_2,\varphi_2)$ is an arrow $\sigma: Z_1 \to Z_2$ such that:

$\varphi_2 = \sigma \circ \varphi_1$

It should be clear that this is exactly the same information encoded by the commutative diagram, so we can think of arrows in our category as these diagrams.

To verify this is a bona-fide category, we need to show 2 things:

1. We have identity arrows.
2. Arrows are composable.

It should be clear that the arrow $1_Z$ serves as an identity arrow for the pair $(Z,\varphi)$. To see that arrows are composable, we join two commutative diagrams along "the matching arrow in the middle".

What Aluffi is doing here, is easing you into the notion of a co-slice category, which in turn is a special case of a "comma category". In this case, we can think of arrows as functions $\tilde{\varphi}:A/\sim \to Z$, and our category is $(A/\sim\downarrow\mathbf{Set})$, or: "the sets under $A/\sim$".

Now what would it mean for an object $(I,f)$ to be initial in this category? It would mean there is a UNIQUE arrow $(I,f) \to (Z,\varphi)$. Remember, $(I,f)$ really means an arrow:

$f:A \to I$ such that $f(a') = f(a)$ for all $a' \sim a$.

Now if: $I = A/\sim$ and $f = \pi$, where $\pi(a) = [a]$ (the equivalence class of $a$), by the definition of ARROWS in our category, we have that an arrow from $(A/\sim,\pi) \to (Z,\varphi)$ is a function:

$g:A/\sim \to Z$ such that $g \circ \pi = \varphi$.

It should (hopefully) be clear that we HAVE to have $g = \tilde{\varphi}$ where:

$\tilde{\varphi}([a]) = \varphi(a)$, for all $a \in A$, by commutativity of the relevant diagram.

In other words, we can induce a well-defined function $\tilde{\varphi}$ for any function $\varphi:A \to Z$ that respects $\sim$ by factoring $\varphi$ through $\pi$.

If you want to "bring this down to earth", suppose that $A = \Bbb Z$, and that we require that all our maps be group homomorphisms. Define $\sim$ by:

$k \sim m \iff k - m \in n\Bbb Z$.

Then "reducing mod $n$" is universal among all group homomorphisms out of the integers that annihilate the multiples of $n$.

Thanks Deveno ... just working through your post carefully now ...

Mind you, I had (somehow) convinced myself that there was only one function/morphism $$ \phi \ : \ A \to Z $$ such that $$ a' \sim a'' \Longrightarrow \phi (a') = \phi (a'') $$ ... BUT as soon as I opened the post and read:

"It should be clear that even if we talk about the same "target" (co-domain) $Z$, we might get several such functions, for example any CONSTANT function $A \to Z$ will respect the equivalence $\sim$."

I realized this notion of mine was a delusion!

I was perhaps thinking of the canonical projection - but then I am thinking even that may have a number of manifestations.

Peter

EDIT: Now not sure about what I said above about canonical projection ... still thinking

 

[Deveno]

What is being discussed here, really, is a consequence of the Fundamental Isomorphism Theorem for Sets:

If $f:A \to B$ is any function, there exists a bijection between $\text{Im }f$ and $A/\sim_f$ where:

$a_1 \sim_f a_2 \iff f(a_1) = f(a_2)$.

I leave it to you to verify that $\sim_f$ is a bona fide equivalence relation.

The bijection in question is:

$\tilde{f}:A/\sim_f \to \text{Im }f$ given by: $\tilde{f}([a]_f) = f(a)$

The first thing to do, of course, is verify $\tilde{f}$ is well-defined.

To this end, we must show if $[a]_f = [a']_f$ (that is $a \sim_f a'$), that $f(a) = f(a')$.

But this is obvious from the definition of $\sim_f$.

To show $\tilde{f}$ is injective, suppose $\tilde{f}([a_1]_f) = \tilde{f}([a_2]_f)$.

This means, $f(a_1) = f(a_2)$, whence $a_1 \sim_f a_2$ and thus $[a_1]_f = [a_2]_f$.

To show $\tilde{f}$ is surjective, suppose $b \in \text{Im }f$. This means that there exists some $a \in A$

with $f(a) = b$. So we have $\tilde{f}[a]_f = f(a) = b$, thus $\tilde{f}$ is surjective.

Let me give you an example to show how this really works:

Suppose $f: \Bbb R \to \Bbb R$ is the squaring function: $f(x) = x^2$.

Then $[x] = \{x,-x\} = \{y \in \Bbb R: y^2 = x^2\}$

Since the squaring function is injective on the non-negative reals, we can denote the equivalence class of $x$ as $|x|$, and our bijection is thus:

$|x| \leftrightarrow x^2$ from $\Bbb R_0^+ \to \Bbb R_0^+$

This is why when we "unsquare" (take a square root) we sometimes get "extraneous answers", because we don't know if it's $x$ or $-x$ we're looking for.

 

[Math Amateur]

What is being discussed here, really, is a consequence of the Fundamental Isomorphism Theorem for Sets:

If $f:A \to B$ is any function, there exists a bijection between $\text{Im }f$ and $A/\sim_f$ where:

$a_1 \sim_f a_2 \iff f(a_1) = f(a_2)$.

I leave it to you to verify that $\sim_f$ is a bona fide equivalence relation.

The bijection in question is:

$\tilde{f}:A/\sim_f \to \text{Im }f$ given by: $\tilde{f}([a]_f) = f(a)$

The first thing to do, of course, is verify $\tilde{f}$ is well-defined.

To this end, we must show if $[a]_f = [a']_f$ (that is $a \sim_f a'$), that $f(a) = f(a')$.

But this is obvious from the definition of $\sim_f$.

To show $\tilde{f}$ is injective, suppose $\tilde{f}([a_1]_f) = \tilde{f}([a_2]_f)$.

This means, $f(a_1) = f(a_2)$, whence $a_1 \sim_f a_2$ and thus $[a_1]_f = [a_2]_f$.

To show $\tilde{f}$ is surjective, suppose $b \in \text{Im }f$. This means that there exists some $a \in A$

with $f(a) = b$. So we have $\tilde{f}[a]_f = f(a) = b$, thus $\tilde{f}$ is surjective.

Let me give you an example to show how this really works:

Suppose $f: \Bbb R \to \Bbb R$ is the squaring function: $f(x) = x^2$.

Then $[x] = \{x,-x\} = \{y \in \Bbb R: y^2 = x^2\}$

Since the squaring function is injective on the non-negative reals, we can denote the equivalence class of $x$ as $|x|$, and our bijection is thus:

$|x| \leftrightarrow x^2$ from $\Bbb R_0^+ \to \Bbb R_0^+$

This is why when we "unsquare" (take a square root) we sometimes get "extraneous answers", because we don't know if it's $x$ or $-x$ we're looking for.

Thanks Deveno ... I really appreciate your help in these matters ... especially as I am finding category theory quite abstract and new ... but I am beginning to see it has great importance for algebra ...

I will be reflecting carefully on what you have said in both your posts above ... and then will work on in Paolo Aluffi's category oriented introduction to algebra ...

Thanks again,

Peter

 

[Math Amateur]

First let's be clear how we are constructing the category Aluffi means:

We start with a set $A$ and an equivalence relation $\sim$ on $A$.

Now the OBJECTS of our category, are ARROWS of $\mathbf{Set}$, in fact the objects under consideration all have a common source ("domain"), the set $A$. Since the objects all have source $A$, we can think of them as a pair: $(Z,\varphi)$.

But not just "any" functions $\varphi:A \to Z$, we only want the functions that induce "well-defined" functions on the set of equivalence classes: $A/\sim$ (these functions are said to RESPECT the equivalence).

It should be clear that even if we talk about the same "target" (co-domain) $Z$, we might get several such functions, for example any CONSTANT function $A \to Z$ will respect the equivalence $\sim$.

I also want to make it clear that we are considering the equivalence relation $\sim$ as FIXED, just like the set $A$.

OK, so we now what the OBJECTS of our category are, what are the ARROWS?

An arrow from $(Z_1,\varphi_1)$ to $(Z_2,\varphi_2)$ is an arrow $\sigma: Z_1 \to Z_2$ such that:

$\varphi_2 = \sigma \circ \varphi_1$

It should be clear that this is exactly the same information encoded by the commutative diagram, so we can think of arrows in our category as these diagrams.

To verify this is a bona-fide category, we need to show 2 things:

1. We have identity arrows.
2. Arrows are composable.

It should be clear that the arrow $1_Z$ serves as an identity arrow for the pair $(Z,\varphi)$. To see that arrows are composable, we join two commutative diagrams along "the matching arrow in the middle".

What Aluffi is doing here, is easing you into the notion of a co-slice category, which in turn is a special case of a "comma category". In this case, we can think of arrows as functions $\tilde{\varphi}:A/\sim \to Z$, and our category is $(A/\sim\downarrow\mathbf{Set})$, or: "the sets under $A/\sim$".

Now what would it mean for an object $(I,f)$ to be initial in this category? It would mean there is a UNIQUE arrow $(I,f) \to (Z,\varphi)$. Remember, $(I,f)$ really means an arrow:

$f:A \to I$ such that $f(a') = f(a)$ for all $a' \sim a$.

Now if: $I = A/\sim$ and $f = \pi$, where $\pi(a) = [a]$ (the equivalence class of $a$), by the definition of ARROWS in our category, we have that an arrow from $(A/\sim,\pi) \to (Z,\varphi)$ is a function:

$g:A/\sim \to Z$ such that $g \circ \pi = \varphi$.

It should (hopefully) be clear that we HAVE to have $g = \tilde{\varphi}$ where:

$\tilde{\varphi}([a]) = \varphi(a)$, for all $a \in A$, by commutativity of the relevant diagram.

In other words, we can induce a well-defined function $\tilde{\varphi}$ for any function $\varphi:A \to Z$ that respects $\sim$ by factoring $\varphi$ through $\pi$.

If you want to "bring this down to earth", suppose that $A = \Bbb Z$, and that we require that all our maps be group homomorphisms. Define $\sim$ by:

$k \sim m \iff k - m \in n\Bbb Z$.

Then "reducing mod $n$" is universal among all group homomorphisms out of the integers that annihilate the multiples of $n$.

Thanks again, Deveno ...

Just a couple of minor issues ...Question/Issue 1.

You write:

"What Aluffi is doing here, is easing you into the notion of a co-slice category, which in turn is a special case of a "comma category". In this case, we can think of arrows as functions $\tilde{\varphi}:A/\sim \to Z$, and our category is $(A/\sim\downarrow\mathbf{Set})$, or: "the sets under $A/\sim$". "

What exactly do you mean (notationally and conceptually) by " ... our category is $(A/\sim\downarrow\mathbf{Set})$" What does the down arrow signify? By the "the sets under $A/\sim$" do you mean the sets making up the partition of A by the equivalence relation?

Question/Issue 2.

You write:

"Now what would it mean for an object $(I,f)$ to be initial in this category? It would mean there is a UNIQUE arrow $(I,f) \to (Z,\varphi)$. ... ... "

Where in your argument do you demonstrate the uniqueness of the arrow $(I,f) \to (Z,\varphi)$ ... that is that there is exactly one such arrow ... can you explicitly point to where in your argument this is achieved ... it is possibly obvious ... but I am not sure on this matter ,,, hope you can help ... ...

Peter

 

[Deveno]

To explain what I mean, I'd have to define a comma category. It's a long explanation, though, and you may not understand it just yet.

Suppose we have 3 categories $\mathcal{A},\mathcal{B},\mathcal{C}$.

Furthermore, suppose we have two functors:

$\mathcal{A}\stackrel{F}{\to}\mathcal{C}\stackrel{G}{\leftarrow}\mathcal{B}$

We can make the comma category $(F\downarrow G)$ as follows:

The objects are triples $(A,B,f)$ where $A$ is an object of $\mathcal{A},\ B$ is an object of $\mathcal{B}$, and $f \in \text{Hom}_{\mathcal{C}}(F(A),G(B))$ (that is an arrow of $\mathcal{C}, f:F(A) \to G(B)$).

The arrows are commutative squares, or equivalently, an arrow between $(A,B,f)$ and $(A',B',f')$ is a pair of arrows $(g,h)$ where:

$g:A \to A'$
$h:B \to B'$

and:

$f' \circ F(g) = G(h) \circ f$

Now some constructions that come up when you use special functors have their own names. If $\mathcal{A} = \mathcal{C}$ and $F$ is the identity functor of $\mathcal{C}$, and $G$ is a functor:

$G:\mathbf{1} \to \mathcal{C}$ (where $\mathbf{1}$ is "the" category with one object ($\ast$), and one (identity) arrow), then what $G$ does is just pick out an object $C$ of $\mathcal{C}$.

In this case, the triples $(A,\ast,f)$ can be abbreviated $(A,f)$ (the object $C$ being understood), and the commutative squares can be "shrunk" to commutative triangles (since $F$ is the identity functor of $\mathcal{C}$, and we don't need to explicitly include identity arrows in commutative diagrams).

Specifically, the objects $(A,f)$ can be identified as arrows $f:A \to C$, so that an arrow $(A,f) \to (A',f')$ is an arrow:

$g:A \to A'$ such that:

$f = f' \circ g$

In this case $(F\downarrow G)$ is often written $(\mathcal{C}\downarrow C)$ and called the slice category of $\mathcal{C}$ over $C$ or "objects over $C$".

Dually, we could have $\mathcal{B} = \mathcal{C}$ with $G$ the identity functor, and $F$ a functor from $\mathbf{1}$. In this case, our commutative triangles have a common source, and this is the co-slice category $(C\downarrow \mathcal{C})$. In this case our objects are arrows $(A,f)$ where:

$f:C \to A$

so that an arrow $(A,f) \to (A',f')$ is an arrow $g:A \to A'$ such that: $g \circ f = f'$.

You've seen this sort of thing before, when you were looking at modules of $R$-module homomorphisms.

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

To answer your second question:

You can think of it this way- any function $f:A \to Z$ that "factors through $\pi$" (that is, respects the equivalence relation $\sim$ on $A$), induces its own partition of $A$, namely the pre-image sets. Thus the partition induced by $\pi$ is a REFINEMENT of the partition induced by $f$. The partition induced by $\pi$, therefore (which is $\sim$) is the FINEST partition of $A$ that respects $\sim$ (we just don't refine $\sim$ any further!). That is $\sim$ is "minimal among partitions that refine $\sim$".

All this really means is that if $f:A \to Z$ gives the same partition (via pre-images $f^{-1}(z)$) as $\sim$, then $A/\sim$ and $Z$ are "isomorphic sets" that is:

$[a] \leftrightarrow f(a)$ is a bijection between $A/\sim$ and $Z$, and that this bijection is uniquely determined by $f$ (not only is there "a" bijection, but there is a SPECIFIC bijection that yields identical partitions of $A$).

The "uniqueness" then, in what I posted before is that $g:A/\sim \to Z$ HAS TO BE $\tilde{\varphi}$, no OTHER function:

$g: A/\sim \to Z$ will have the property that:

$g \circ \pi = \varphi$

(since $g$ respects $\sim$ it has to be constant on the equivalence classes of $\sim$, and the value it takes on each equivalence class has to be $\varphi(b)$ for every $b \in [a]$).

In group theory, we have the equivalence relation $\sim_H$ given by $g \sim_H g' \iff g^{-1}g' \in H$, for a subgroup $H$ of $G$. This equivalence is used to show that there is just one possible mapping, given $\phi: G \to G'$ (a homomorphism) from $G/\text{ker }\phi \to G'$ that respects $\sim_{\text{ker }\phi}$, namely:

$g\text{ker}\phi \mapsto \phi(g)$

In exactly the same way we say the quotient mapping $G \to G/H$ is universal among group homomorphisms that respect $\sim_H$ (clearly any such homomorphism has to send all of $H$ to the identity, since $e \in H$ and for any homomorphism $\phi:G \to G'$, we have $\phi(e) = e'$). Another way to say this, is that any normal subgroup $H$ of $G$ induces a (group-creating) partition of $G$, and $H$ creates the finest such partition that includes $H$ as one of its pieces.

 

[Math Amateur]

To explain what I mean, I'd have to define a comma category. It's a long explanation, though, and you may not understand it just yet.

Suppose we have 3 categories $\mathcal{A},\mathcal{B},\mathcal{C}$.

Furthermore, suppose we have two functors:

$\mathcal{A}\stackrel{F}{\to}\mathcal{C}\stackrel{G}{\leftarrow}\mathcal{B}$

We can make the comma category $(F\downarrow G)$ as follows:

The objects are triples $(A,B,f)$ where $A$ is an object of $\mathcal{A},\ B$ is an object of $\mathcal{B}$, and $f \in \text{Hom}_{\mathcal{C}}(F(A),G(B))$ (that is an arrow of $\mathcal{C}, f:F(A) \to G(B)$).

The arrows are commutative squares, or equivalently, an arrow between $(A,B,f)$ and $(A',B',f')$ is a pair of arrows $(g,h)$ where:

$g:A \to A'$
$h:B \to B'$

and:

$f' \circ F(g) = G(h) \circ f$

Now some constructions that come up when you use special functors have their own names. If $\mathcal{A} = \mathcal{C}$ and $F$ is the identity functor of $\mathcal{C}$, and $G$ is a functor:

$G:\mathbf{1} \to \mathcal{C}$ (where $\mathbf{1}$ is "the" category with one object ($\ast$), and one (identity) arrow), then what $G$ does is just pick out an object $C$ of $\mathcal{C}$.

In this case, the triples $(A,\ast,f)$ can be abbreviated $(A,f)$ (the object $C$ being understood), and the commutative squares can be "shrunk" to commutative triangles (since $F$ is the identity functor of $\mathcal{C}$, and we don't need to explicitly include identity arrows in commutative diagrams).

Specifically, the objects $(A,f)$ can be identified as arrows $f:A \to C$, so that an arrow $(A,f) \to (A',f')$ is an arrow:

$g:A \to A'$ such that:

$f = f' \circ g$

In this case $(F\downarrow G)$ is often written $(\mathcal{C}\downarrow C)$ and called the slice category of $\mathcal{C}$ over $C$ or "objects over $C$".

Dually, we could have $\mathcal{B} = \mathcal{C}$ with $G$ the identity functor, and $F$ a functor from $\mathbf{1}$. In this case, our commutative triangles have a common source, and this is the co-slice category $(C\downarrow \mathcal{C})$. In this case our objects are arrows $(A,f)$ where:

$f:C \to A$

so that an arrow $(A,f) \to (A',f')$ is an arrow $g:A \to A'$ such that: $g \circ f = f'$.

You've seen this sort of thing before, when you were looking at modules of $R$-module homomorphisms.

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

To answer your second question:

You can think of it this way- any function $f:A \to Z$ that "factors through $\pi$" (that is, respects the equivalence relation $\sim$ on $A$), induces its own partition of $A$, namely the pre-image sets. Thus the partition induced by $\pi$ is a REFINEMENT of the partition induced by $f$. The partition induced by $\pi$, therefore (which is $\sim$) is the FINEST partition of $A$ that respects $\sim$ (we just don't refine $\sim$ any further!). That is $\sim$ is "minimal among partitions that refine $\sim$".

All this really means is that if $f:A \to Z$ gives the same partition (via pre-images $f^{-1}(z)$) as $\sim$, then $A/\sim$ and $Z$ are "isomorphic sets" that is:

$[a] \leftrightarrow f(a)$ is a bijection between $A/\sim$ and $Z$, and that this bijection is uniquely determined by $f$ (not only is there "a" bijection, but there is a SPECIFIC bijection that yields identical partitions of $A$).

The "uniqueness" then, in what I posted before is that $g:A/\sim \to Z$ HAS TO BE $\tilde{\varphi}$, no OTHER function:

$g: A/\sim \to Z$ will have the property that:

$g \circ \pi = \varphi$

(since $g$ respects $\sim$ it has to be constant on the equivalence classes of $\sim$, and the value it takes on each equivalence class has to be $\varphi(b)$ for every $b \in [a]$).

In group theory, we have the equivalence relation $\sim_H$ given by $g \sim_H g' \iff g^{-1}g' \in H$, for a subgroup $H$ of $G$. This equivalence is used to show that there is just one possible mapping, given $\phi: G \to G'$ (a homomorphism) from $G/\text{ker }\phi \to G'$ that respects $\sim_{\text{ker }\phi}$, namely:

$g\text{ker}\phi \mapsto \phi(g)$

In exactly the same way we say the quotient mapping $G \to G/H$ is universal among group homomorphisms that respect $\sim_H$ (clearly any such homomorphism has to send all of $H$ to the identity, since $e \in H$ and for any homomorphism $\phi:G \to G'$, we have $\phi(e) = e'$). Another way to say this, is that any normal subgroup $H$ of $G$ induces a (group-creating) partition of $G$, and $H$ creates the finest such partition that includes $H$ as one of its pieces.

Deveno, thank you for an extremely instructive and helpful post ... I would not be progressing through category theory without such support ...

Reading through your post carefully now and reflecting on what you have written ... Thank you again ...

Peter

 

[Math Amateur]

To explain what I mean, I'd have to define a comma category. It's a long explanation, though, and you may not understand it just yet.

Suppose we have 3 categories $\mathcal{A},\mathcal{B},\mathcal{C}$.

Furthermore, suppose we have two functors:

$\mathcal{A}\stackrel{F}{\to}\mathcal{C}\stackrel{G}{\leftarrow}\mathcal{B}$

We can make the comma category $(F\downarrow G)$ as follows:

The objects are triples $(A,B,f)$ where $A$ is an object of $\mathcal{A},\ B$ is an object of $\mathcal{B}$, and $f \in \text{Hom}_{\mathcal{C}}(F(A),G(B))$ (that is an arrow of $\mathcal{C}, f:F(A) \to G(B)$).

The arrows are commutative squares, or equivalently, an arrow between $(A,B,f)$ and $(A',B',f')$ is a pair of arrows $(g,h)$ where:

$g:A \to A'$
$h:B \to B'$

and:

$f' \circ F(g) = G(h) \circ f$

Now some constructions that come up when you use special functors have their own names. If $\mathcal{A} = \mathcal{C}$ and $F$ is the identity functor of $\mathcal{C}$, and $G$ is a functor:

$G:\mathbf{1} \to \mathcal{C}$ (where $\mathbf{1}$ is "the" category with one object ($\ast$), and one (identity) arrow), then what $G$ does is just pick out an object $C$ of $\mathcal{C}$.

In this case, the triples $(A,\ast,f)$ can be abbreviated $(A,f)$ (the object $C$ being understood), and the commutative squares can be "shrunk" to commutative triangles (since $F$ is the identity functor of $\mathcal{C}$, and we don't need to explicitly include identity arrows in commutative diagrams).

Specifically, the objects $(A,f)$ can be identified as arrows $f:A \to C$, so that an arrow $(A,f) \to (A',f')$ is an arrow:

$g:A \to A'$ such that:

$f = f' \circ g$

In this case $(F\downarrow G)$ is often written $(\mathcal{C}\downarrow C)$ and called the slice category of $\mathcal{C}$ over $C$ or "objects over $C$".

Dually, we could have $\mathcal{B} = \mathcal{C}$ with $G$ the identity functor, and $F$ a functor from $\mathbf{1}$. In this case, our commutative triangles have a common source, and this is the co-slice category $(C\downarrow \mathcal{C})$. In this case our objects are arrows $(A,f)$ where:

$f:C \to A$

so that an arrow $(A,f) \to (A',f')$ is an arrow $g:A \to A'$ such that: $g \circ f = f'$.

You've seen this sort of thing before, when you were looking at modules of $R$-module homomorphisms.

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

To answer your second question:

You can think of it this way- any function $f:A \to Z$ that "factors through $\pi$" (that is, respects the equivalence relation $\sim$ on $A$), induces its own partition of $A$, namely the pre-image sets. Thus the partition induced by $\pi$ is a REFINEMENT of the partition induced by $f$. The partition induced by $\pi$, therefore (which is $\sim$) is the FINEST partition of $A$ that respects $\sim$ (we just don't refine $\sim$ any further!). That is $\sim$ is "minimal among partitions that refine $\sim$".

All this really means is that if $f:A \to Z$ gives the same partition (via pre-images $f^{-1}(z)$) as $\sim$, then $A/\sim$ and $Z$ are "isomorphic sets" that is:

$[a] \leftrightarrow f(a)$ is a bijection between $A/\sim$ and $Z$, and that this bijection is uniquely determined by $f$ (not only is there "a" bijection, but there is a SPECIFIC bijection that yields identical partitions of $A$).

The "uniqueness" then, in what I posted before is that $g:A/\sim \to Z$ HAS TO BE $\tilde{\varphi}$, no OTHER function:

$g: A/\sim \to Z$ will have the property that:

$g \circ \pi = \varphi$

(since $g$ respects $\sim$ it has to be constant on the equivalence classes of $\sim$, and the value it takes on each equivalence class has to be $\varphi(b)$ for every $b \in [a]$).

In group theory, we have the equivalence relation $\sim_H$ given by $g \sim_H g' \iff g^{-1}g' \in H$, for a subgroup $H$ of $G$. This equivalence is used to show that there is just one possible mapping, given $\phi: G \to G'$ (a homomorphism) from $G/\text{ker }\phi \to G'$ that respects $\sim_{\text{ker }\phi}$, namely:

$g\text{ker}\phi \mapsto \phi(g)$

In exactly the same way we say the quotient mapping $G \to G/H$ is universal among group homomorphisms that respect $\sim_H$ (clearly any such homomorphism has to send all of $H$ to the identity, since $e \in H$ and for any homomorphism $\phi:G \to G'$, we have $\phi(e) = e'$). Another way to say this, is that any normal subgroup $H$ of $G$ induces a (group-creating) partition of $G$, and $H$ creates the finest such partition that includes $H$ as one of its pieces.

Thanks again for your help Deveno ... much appreciated.

Now just an issue regarding refinements of partitions ...

You write:

"To answer your second question:

You can think of it this way- any function [FONT=MathJax_Math]f[FONT=MathJax_Main]:[FONT=MathJax_Math]A[FONT=MathJax_Main]→[FONT=MathJax_Math]Z that "factors through [FONT=MathJax_Math]π" (that is, respects the equivalence relation [FONT=MathJax_Main]∼ on [FONT=MathJax_Math]A), induces its own partition of [FONT=MathJax_Math]A, namely the pre-image sets. Thus the partition induced by [FONT=MathJax_Math]π is a REFINEMENT of the partition induced by [FONT=MathJax_Math]f. The partition induced by [FONT=MathJax_Math]π, therefore (which is [FONT=MathJax_Main]∼) is the FINEST partition of [FONT=MathJax_Math]A that respects [FONT=MathJax_Main]∼ (we just don't refine [FONT=MathJax_Main]∼ any further!). That is [FONT=MathJax_Main]∼ is "minimal among partitions that refine [FONT=MathJax_Main]∼". ... ... "

Now I have an intuitive sense of what constitutes the refinement of a partition ...BUT ... wish to check with you ... just to be sure ...

Consider Figures 1 and 2 as follows:View attachment 2630Figure 1 shows a partition of set $$A$$ by the equivalence relation $$ \sim $$.

$$ \sim $$ partitions $$A$$ into $$ [a_1]_\sim , [a_2]_\sim \text{ and } [a_3]_\sim $$.

What I take to be a refinement of the partition brought about by $$ \sim $$ is shown in Figure 2.

Figure 2 shows a partition that (in your words) "respects" the boundaries of $$ \sim $$ but also breaks the partition down into a finer partition which comprises $$ [a_1]_{\sim_R} , [a_3]_{\sim_R} , [a_5]_{\sim_R} , [a_7]_{\sim_R}, [a_8]_{\sim_R} $$

Can you please confirm (or otherwise) that my notion of a refinement is correct?

Further, consider Figure 3 as follows:View attachment 2631

Figure 3 above shows a morphism $$ \sigma \ : \ A \to Z $$

The morphism $$ \sigma $$ "respects" the partition enacted by $$ \sim $$ since for $$ \sigma $$ we have that $$ m' = m'' \Longrightarrow \sigma (m') = \sigma (m'') $$ but the pre-images give a coarser partition since we essentially have the following:

$$ {\sigma}^{-1} ( z_1 ) = [a_1]_\sim \cup \ [a_5]_\sim $$

and

$$ {\sigma}^{-1} ( z_2 ) = [a_7]_\sim $$

Can you confirm that my reasoning is correct?

Peter

 

[Deveno]

That is exactly right. The fineness of the partition induced by a map $f$ on the domain (via pre-images) gives some sort of idea of "how injective" $f$ is. For example, a constant function yields a VERY coarse partition, and an injective function yields the finest partition possible of the domain into singletons.

It's somewhat helpful to think of maps this way, because it generalizes well to understanding topology, where we are VERY concerned with what pre-images look like. The basic idea of topology is that "points" (individual set elements) are replaced by "neighborhoods" (essentially somewhat fuzzy blobs around the points).

The objection is often levied against category theory that it is devoid of content. This is "somewhat" true. Category theory is more of a "narrative structure" than a "thing".

To continue the analogy further, when a playwright is writing a play, the actual names of the characters in the play don't really matter that much. What matters is the interactions of the characters with each other and the audience. A murder mystery may feature a detective named Elwood in the 19th century, or someone named Charlene in the 21st. Plot devices that are "essentially the same" are employed over and over again: the surprise twist, the red herring, the dramatic confession.

In different realms of mathematics, certain "canonical" constructions appear over and over again: the product, sub-objects, quotient objects, invariant decompositions. The fundamental isomorphism theorems can be stated for several categories, and in each case the proof of them follows the same script.

Sometimes, of course, there are unique facets to a particular category. But even then, we often have "some transfer". Here is an example:

Given a commutative ring with unity, $R$, one can form the group of units of $R,\ U(R)$. It turns out that any ring homomorphsim:

$\phi:R \to S$

induces an homomorphism of abelian groups:

$U(\phi): U(R) \to U(S)$.

This is an example of a functor: $\mathbf{CRing} \to \mathbf{AbGrp}$.

The thing about this construction is: we don't need to know ANYTHING about the details of the ring $R$ to establish this, we do it "in the same way" for every commutative ring with identity, it's "categorical".

This turns out to be very useful, even in "particular cases". The particular functor above pays immense dividends in number theory, where information about the group of units mod $n$, can be "lifted" to information about integers, and then used to solve things like Diophantine equations.

Sometimes, there's information hidden in "the obvious" that just needs the right "framing" to make it stand out.

 

[Math Amateur]

That is exactly right. The fineness of the partition induced by a map $f$ on the domain (via pre-images) gives some sort of idea of "how injective" $f$ is. For example, a constant function yields a VERY coarse partition, and an injective function yields the finest partition possible of the domain into singletons.

It's somewhat helpful to think of maps this way, because it generalizes well to understanding topology, where we are VERY concerned with what pre-images look like. The basic idea of topology is that "points" (individual set elements) are replaced by "neighborhoods" (essentially somewhat fuzzy blobs around the points).

The objection is often levied against category theory that it is devoid of content. This is "somewhat" true. Category theory is more of a "narrative structure" than a "thing".

To continue the analogy further, when a playwright is writing a play, the actual names of the characters in the play don't really matter that much. What matters is the interactions of the characters with each other and the audience. A murder mystery may feature a detective named Elwood in the 19th century, or someone named Charlene in the 21st. Plot devices that are "essentially the same" are employed over and over again: the surprise twist, the red herring, the dramatic confession.

In different realms of mathematics, certain "canonical" constructions appear over and over again: the product, sub-objects, quotient objects, invariant decompositions. The fundamental isomorphism theorems can be stated for several categories, and in each case the proof of them follows the same script.

Sometimes, of course, there are unique facets to a particular category. But even then, we often have "some transfer". Here is an example:

Given a commutative ring with unity, $R$, one can form the group of units of $R,\ U(R)$. It turns out that any ring homomorphsim:

$\phi:R \to S$

induces an homomorphism of abelian groups:

$U(\phi): U(R) \to U(S)$.

This is an example of a functor: $\mathbf{CRing} \to \mathbf{AbGrp}$.

The thing about this construction is: we don't need to know ANYTHING about the details of the ring $R$ to establish this, we do it "in the same way" for every commutative ring with identity, it's "categorical".

This turns out to be very useful, even in "particular cases". The particular functor above pays immense dividends in number theory, where information about the group of units mod $n$, can be "lifted" to information about integers, and then used to solve things like Diophantine equations.

Sometimes, there's information hidden in "the obvious" that just needs the right "framing" to make it stand out.

Thanks Deveno ... Very helpful in giving me the broad picture ...

Peter

 

[Math Amateur]

To explain what I mean, I'd have to define a comma category. It's a long explanation, though, and you may not understand it just yet.

Suppose we have 3 categories $\mathcal{A},\mathcal{B},\mathcal{C}$.

Furthermore, suppose we have two functors:

$\mathcal{A}\stackrel{F}{\to}\mathcal{C}\stackrel{G}{\leftarrow}\mathcal{B}$

We can make the comma category $(F\downarrow G)$ as follows:

The objects are triples $(A,B,f)$ where $A$ is an object of $\mathcal{A},\ B$ is an object of $\mathcal{B}$, and $f \in \text{Hom}_{\mathcal{C}}(F(A),G(B))$ (that is an arrow of $\mathcal{C}, f:F(A) \to G(B)$).

The arrows are commutative squares, or equivalently, an arrow between $(A,B,f)$ and $(A',B',f')$ is a pair of arrows $(g,h)$ where:

$g:A \to A'$
$h:B \to B'$

and:

$f' \circ F(g) = G(h) \circ f$

Now some constructions that come up when you use special functors have their own names. If $\mathcal{A} = \mathcal{C}$ and $F$ is the identity functor of $\mathcal{C}$, and $G$ is a functor:

$G:\mathbf{1} \to \mathcal{C}$ (where $\mathbf{1}$ is "the" category with one object ($\ast$), and one (identity) arrow), then what $G$ does is just pick out an object $C$ of $\mathcal{C}$.

In this case, the triples $(A,\ast,f)$ can be abbreviated $(A,f)$ (the object $C$ being understood), and the commutative squares can be "shrunk" to commutative triangles (since $F$ is the identity functor of $\mathcal{C}$, and we don't need to explicitly include identity arrows in commutative diagrams).

Specifically, the objects $(A,f)$ can be identified as arrows $f:A \to C$, so that an arrow $(A,f) \to (A',f')$ is an arrow:

$g:A \to A'$ such that:

$f = f' \circ g$

In this case $(F\downarrow G)$ is often written $(\mathcal{C}\downarrow C)$ and called the slice category of $\mathcal{C}$ over $C$ or "objects over $C$".

Dually, we could have $\mathcal{B} = \mathcal{C}$ with $G$ the identity functor, and $F$ a functor from $\mathbf{1}$. In this case, our commutative triangles have a common source, and this is the co-slice category $(C\downarrow \mathcal{C})$. In this case our objects are arrows $(A,f)$ where:

$f:C \to A$

so that an arrow $(A,f) \to (A',f')$ is an arrow $g:A \to A'$ such that: $g \circ f = f'$.

You've seen this sort of thing before, when you were looking at modules of $R$-module homomorphisms.

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

To answer your second question:

You can think of it this way- any function $f:A \to Z$ that "factors through $\pi$" (that is, respects the equivalence relation $\sim$ on $A$), induces its own partition of $A$, namely the pre-image sets. Thus the partition induced by $\pi$ is a REFINEMENT of the partition induced by $f$. The partition induced by $\pi$, therefore (which is $\sim$) is the FINEST partition of $A$ that respects $\sim$ (we just don't refine $\sim$ any further!). That is $\sim$ is "minimal among partitions that refine $\sim$".

All this really means is that if $f:A \to Z$ gives the same partition (via pre-images $f^{-1}(z)$) as $\sim$, then $A/\sim$ and $Z$ are "isomorphic sets" that is:

$[a] \leftrightarrow f(a)$ is a bijection between $A/\sim$ and $Z$, and that this bijection is uniquely determined by $f$ (not only is there "a" bijection, but there is a SPECIFIC bijection that yields identical partitions of $A$).

The "uniqueness" then, in what I posted before is that $g:A/\sim \to Z$ HAS TO BE $\tilde{\varphi}$, no OTHER function:

$g: A/\sim \to Z$ will have the property that:

$g \circ \pi = \varphi$

(since $g$ respects $\sim$ it has to be constant on the equivalence classes of $\sim$, and the value it takes on each equivalence class has to be $\varphi(b)$ for every $b \in [a]$).

In group theory, we have the equivalence relation $\sim_H$ given by $g \sim_H g' \iff g^{-1}g' \in H$, for a subgroup $H$ of $G$. This equivalence is used to show that there is just one possible mapping, given $\phi: G \to G'$ (a homomorphism) from $G/\text{ker }\phi \to G'$ that respects $\sim_{\text{ker }\phi}$, namely:

$g\text{ker}\phi \mapsto \phi(g)$

In exactly the same way we say the quotient mapping $G \to G/H$ is universal among group homomorphisms that respect $\sim_H$ (clearly any such homomorphism has to send all of $H$ to the identity, since $e \in H$ and for any homomorphism $\phi:G \to G'$, we have $\phi(e) = e'$). Another way to say this, is that any normal subgroup $H$ of $G$ induces a (group-creating) partition of $G$, and $H$ creates the finest such partition that includes $H$ as one of its pieces.

Hi Deveno ... ... I thought I would follow your advice, rendered on a number of occasions, to use small examples in order to get a better sense and understanding of a notion, proposition, theorem or the like.

In this case, I am seeking to fully understand Aluffi's Claim 5.5 on page 34. Claim 5.5 asserts that $$ ( \pi, A/\sim ) $$ is an initial object of the category of functions $$ \phi \ : \ A \to Z $$ (where Z is any set and A is a particular given set) satisfying the property that $$ a' \sim a'' \Longrightarrow \phi (a') = \phi (a'') $$.

To show that $$ ( \pi, A/ \sim ) $$is an initial object we need to consider any (an arbitrary) object $$ ( \phi, Z) $$ and then show that there exists a UNIQUE (exactly one) morphism $$ \overline{ \phi } \ : \ ( \pi, A/ \sim ) \to ( \phi, Z) $$ ... ... that is there is a UNIQUE commutative diagram as shown in Figure 1 below:View attachment 2645In other words, there is a UNIQUE (exactly one) morphism $$ \overline{ \phi } $$ making the diagram in Figure 1 commute.

Now, consider the following set A shown as partitioned by an equivalence relation $$ \sim $$.
View attachment 2646

Now consider the functions/mappings $$ \pi \ : \ a \to [a]_\sim $$ and $$ \phi \ : \ A \to Z $$ that are shown in Figure 3 below.View attachment 2647

Now we have the following mappings or functions as follows:


(1)

$$ \pi $$ where

$$ \pi(a_1) = \pi(a_2) = \pi(a_3) = \pi(a_4) = [a_1]_\sim $$

and

$$ \pi(a_5) = \pi(a_6) = [a_5]_\sim $$

and

$$ \pi(a_7) = \pi(a_8) = \pi(a_9) = [a_7]_\sim $$


(2)

$$ \phi $$ where

$$ \phi(a_1) = \phi(a_2) = \phi(a_3) = \phi(a_4) = z_1 $$

and

$$ \phi(a_5) = \phi(a_6) = z_1 $$

and

$$ \phi(a_7) = \phi(a_8) = \phi(a_9) = z_2 $$

*** Note that given these functional assignments that $$ \phi $$ respects the partition enacted by $$ \sim $$

*** Note also that there are other possibilities we could have chosen for a map that respects the particular partition.Now, given $$ A, A/ \sim, Z $$ and the morphisms $$ \pi , \phi $$ as shown, there is exactly and only one possibility for $$ \overline{ \phi } $$ given that we must have $$ \pi \circ \overline{ \phi } = \phi $$ and that is the following assignments:

$$ \overline{ \phi } ( [a_1]_\sim ) = z_1 $$

and

$$ \overline{ \phi } ( [a_2]_\sim ) = z_1 $$

and

$$ \overline{ \phi } ( [a_3]_\sim ) = z_2 $$

Thus, since there is only one possibility for $$ \overline{ \phi } $$, it is, by definition, UNIQUE, given $$ \pi $$ and the particular $$ \phi $$ chosen.

For a different $$ \phi $$ that also respects $$ \sim $$, we would have a different, but UNIQUE $$ \overline{ \phi } $$.

Can someone please confirm that my analysis regarding the above example is correct?

Peter
***EDIT***

One further question I have related to the term "universal properties" is as follows:

In what way does the existence of a unique morphism $$ \overline{ \phi }$$ amount so a universal property? In what way is it "universal"? Is it "universal" because a unique $$ \overline{ \phi }$$ exists for every object $$ ( \phi , Z ) $$ in the category.

As MHB members will gather from this question I am having problems getting a real sense of the idea of "universal properties" in category theory!

So any help in this will be much appreciated!Peter

 

[Deveno]

Yes, that looks right.

Let's make another example:

Suppose $A = \Bbb Z$, and $\pi$ is the mapping $k \mapsto [k]_{12}$ (the equivalence class of $k$ modulo 12, so our equivalence relation is "equivalent mod 12").

Suppose as well that $Z = \Bbb Z/4\Bbb Z$ and that $\phi:\Bbb Z \to \Bbb Z/4\Bbb Z$ is the mapping $\phi(k) = [k]_4$.

We need to find a UNIQUE mapping $\overline{\phi}:\Bbb Z/12\Bbb Z \to \Bbb Z/4\Bbb Z$ such that:

$\phi = \overline{\phi} \circ \pi$.

Before we start, we should verify that $\phi$ respects the equivalence "mod 12". This means showing:

If $k \equiv m \text{ (mod }12)$ then $k \equiv m \text{ (mod }4)$.

Now the first part means that $k - m = 12t$ for some $t \in \Bbb Z$.

Thus $k - m = 4(3t)$ and since $3t \in \Bbb Z$ whenever $t$ is, $\phi$ is a function that respects "mod 12".

(you can see that this is precisely because mod 12 refines mod 4, which happens because 4 divides 12).

So what could $\overline{\phi}$ be?

Let's follow the integer 1 through our commutative diagram. on one path, we get:

$1 \to [1]_4$. on the other path, we get:

$1 \to [1]_{12} \to [1]_4$ (since the diagram commutes).

Nothing really exciting happens for 0,2, or 3, so let's skip ahead to what happens to 4:

$4 \to [0]_4$
$4 \to [4]_{12} \to [0]_4$ <--we reduce what is "inside the brackets" mod 4.

It looks like what $\overline{\phi}$ HAS to be is $\overline{\phi}([k]_{12}) = [k]_4$

Let's test this theory: by the division algorithm, we can write any positive integer as:

$k = 12q + r$ and

$r = 4q' + r'$.

This means we can write $k = 12q + 4q' + r'$.

Now $\phi(k) = [r']_4$, and $\pi(k) = [r]_{12} = [4q' + r']_{12}$.

Since our diagram commutes, we have:

$\overline{\phi}([r]_{12}) = [r']_4$, which is clearly $r$ (mod 4).

(I leave it to you to investigate how we handle non-positive integers).

So, for example, to compute $\overline{\phi}([647]_{12})$, we first reduce 647 mod 12, which gives us 11.

We then reduce 11 mod 4, which gives us 3, so:

$\overline{\phi}([647]_{12}) = [3]_4$.

This should be the same as 647 mod 4, which it is.

The function $\overline{\phi}$ in these discussions is often called the function induced on $A/\sim$ by $\phi$. Checking that $\overline{\phi}$ is "well-defined" is equivalent to saying $\phi$ respects $\sim$.**************************

The upshot of this is that "quotient sets" (set partitioned by an equivalence) and "surjective functions" are really the same thing in disguise. For every equivalence relation $\sim$ we get the surjective function:

$\pi:A \to A/\sim$ with $\pi(a) = [a]_{\sim}$

and for every surjective function:

$f: A \to f(A)$ we get the equivalence relation:

$\sim_f: a_1 \sim_f a_2 \iff f(a_1) = f(a_2)$

(note that $\sim$ in the first example IS $\sim_{\pi}$).

So why prefer a "surjective function" definition of equivalence class? Because we don't have to look at "representatives" (elements of $A$), we can check for surjectivity "globally" by finding a right-inverse.

The universal property of the "quotient function" $\pi$ means that $\pi$ is CANONICAL, any function out of the set of equivalences classes of $\sim$ has to "go through $\pi$" first.

I want to stress that the function $\overline{\phi}$ is "almost invisible", in actual practice, one usually just shows that it is well-defined, and then proceeds to work with $\phi$. This "decomposition" of a function into an equivalence, and a bijection is almost trivial. What is NOT trivial, is that it happens to be THE SAME PROOF, when:

"set" is replaced by "group"
"equivalence" is replaced by "modulo a normal subgroup"
"function" is replaced by "group homomorphism"

which means that quotient groups satisfy the SAME universal property, in a different category. And perhaps you can see why this might be more important than it seems at first glance: all of a sudden quotient groups and normal subgroups aren't quite so arbitrary, they come about naturally as soon as you have the notion of homomorphism.

Perhaps similar things might happen in MANY categories, in which case, universal properties might serve as a way of identifying the important CONCEPTS to look for in a given structure.

Try to formulate what the universal property of a quotient group might be. Your answer should somehow involve the canonical projection: $\pi: G \to G/N$. What would "respecting $\sim_N$" mean? Doing this might help you with the "set version".

 

[Math Amateur]

Yes, that looks right.

Let's make another example:

Suppose $A = \Bbb Z$, and $\pi$ is the mapping $k \mapsto [k]_{12}$ (the equivalence class of $k$ modulo 12, so our equivalence relation is "equivalent mod 12").

Suppose as well that $Z = \Bbb Z/4\Bbb Z$ and that $\phi:\Bbb Z \to \Bbb Z/4\Bbb Z$ is the mapping $\phi(k) = [k]_4$.

We need to find a UNIQUE mapping $\overline{\phi}:\Bbb Z/12\Bbb Z \to \Bbb Z/4\Bbb Z$ such that:

$\phi = \overline{\phi} \circ \pi$.

Before we start, we should verify that $\phi$ respects the equivalence "mod 12". This means showing:

If $k \equiv m \text{ (mod }12)$ then $k \equiv m \text{ (mod }4)$.

Now the first part means that $k - m = 12t$ for some $t \in \Bbb Z$.

Thus $k - m = 4(3t)$ and since $3t \in \Bbb Z$ whenever $t$ is, $\phi$ is a function that respects "mod 12".

(you can see that this is precisely because mod 12 refines mod 4, which happens because 4 divides 12).

So what could $\overline{\phi}$ be?

Let's follow the integer 1 through our commutative diagram. on one path, we get:

$1 \to [1]_4$. on the other path, we get:

$1 \to [1]_{12} \to [1]_4$ (since the diagram commutes).

Nothing really exciting happens for 0,2, or 3, so let's skip ahead to what happens to 4:

$4 \to [0]_4$
$4 \to [4]_{12} \to [0]_4$ <--we reduce what is "inside the brackets" mod 4.

It looks like what $\overline{\phi}$ HAS to be is $\overline{\phi}([k]_{12}) = [k]_4$

Let's test this theory: by the division algorithm, we can write any positive integer as:

$k = 12q + r$ and

$r = 4q' + r'$.

This means we can write $k = 12q + 4q' + r'$.

Now $\phi(k) = [r']_4$, and $\pi(k) = [r]_{12} = [4q' + r']_{12}$.

Since our diagram commutes, we have:

$\overline{\phi}([r]_{12}) = [r']_4$, which is clearly $r$ (mod 4).

(I leave it to you to investigate how we handle non-positive integers).

So, for example, to compute $\overline{\phi}([647]_{12})$, we first reduce 647 mod 12, which gives us 11.

We then reduce 11 mod 4, which gives us 3, so:

$\overline{\phi}([647]_{12}) = [3]_4$.

This should be the same as 647 mod 4, which it is.

The function $\overline{\phi}$ in these discussions is often called the function induced on $A/\sim$ by $\phi$. Checking that $\overline{\phi}$ is "well-defined" is equivalent to saying $\phi$ respects $\sim$.**************************

The upshot of this is that "quotient sets" (set partitioned by an equivalence) and "surjective functions" are really the same thing in disguise. For every equivalence relation $\sim$ we get the surjective function:

$\pi:A \to A/\sim$ with $\pi(a) = [a]_{\sim}$

and for every surjective function:

$f: A \to f(A)$ we get the equivalence relation:

$\sim_f: a_1 \sim_f a_2 \iff f(a_1) = f(a_2)$

(note that $\sim$ in the first example IS $\sim_{\pi}$).

So why prefer a "surjective function" definition of equivalence class? Because we don't have to look at "representatives" (elements of $A$), we can check for surjectivity "globally" by finding a right-inverse.

The universal property of the "quotient function" $\pi$ means that $\pi$ is CANONICAL, any function out of the set of equivalences classes of $\sim$ has to "go through $\pi$" first.

I want to stress that the function $\overline{\phi}$ is "almost invisible", in actual practice, one usually just shows that it is well-defined, and then proceeds to work with $\phi$. This "decomposition" of a function into an equivalence, and a bijection is almost trivial. What is NOT trivial, is that it happens to be THE SAME PROOF, when:

"set" is replaced by "group"
"equivalence" is replaced by "modulo a normal subgroup"
"function" is replaced by "group homomorphism"

which means that quotient groups satisfy the SAME universal property, in a different category. And perhaps you can see why this might be more important than it seems at first glance: all of a sudden quotient groups and normal subgroups aren't quite so arbitrary, they come about naturally as soon as you have the notion of homomorphism.

Perhaps similar things might happen in MANY categories, in which case, universal properties might serve as a way of identifying the important CONCEPTS to look for in a given structure.

Try to formulate what the universal property of a quotient group might be. Your answer should somehow involve the canonical projection: $\pi: G \to G/N$. What would "respecting $\sim_N$" mean? Doing this might help you with the "set version".

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

Peter

***EDIT***

Thanks in no small part to your help I may soon be able to go back to Paul E Bland's book: Rings and Their Modules. I had to leave working on that book because (especially when it came to the section on Direct Products and Sums, Bland took a category oriented approach.

Without some understanding of category theory it is extremely difficult to understand the motivation for the definition of various constructions such as direct products and sums ... the definitions seem weird and abstract until you understand what the author is up to!