De-mystifying universal mapping properties: an example-quotient groups.

[Deveno]

Often, in the study of algebraic objects certain things (like tensor products) are often defined primarily in terms of an universal mapping property. When one is used to "concrete objects" one can calculate with, this often comes as a shock to the system. One feels as if one is spinning something out of nothing, and has the feeling of "waiting for the real 'stuff' to show up."

Hopefully, in this small example, I will demonstrate there is nothing to be afraid of in this type of construction, and that viewing algebraic objects this way can often be clearer that the "concrete" presentation. The example I have chosen is the humble quotient group, often a major stumbling block in terms of comprehension for those just beginning a survey of abstract algebra.

In a spirit of sly perversity, I will begin with the "abstract" definition, and derive some commonly proven theorems, before delving into the "guts" of the beast, and showing what we are talking about is "the same" as the typical introduction to quotient groups.

We have in front of us a lot of ground to cover, and since we will be focusing on mappings, in general, I will first start with some observations about things I hope most readers are familiar with: ordinary functions.

A function $f: B \to C$ is said to be one-to-one or injective, if there is no "condensing" of $B$ that takes place, each separate element of $B$ is mapped to a unique element of $C$ in such a way that no two elements of $B$ are taken by $f$ to the same element of $C$. This is usually presented by the statement:

$f$ is injective if: $f(b_1) = f(b_2) \implies b_1 = b_2$.

We will use this as our "concrete definition" (for now). Soon, we will replace it with:

$f$ is injective $\iff$ for any two functions $g_1,g_2:A \to B$ (with $A$ any set):

$f\circ g_1 = f\circ g_2 \implies g_1 = g_2$.

Of course, we have to PROVE these two definitions are equivalent:

Definition 2 $\implies$ Definition 1:

Since Definition 2 applies to any set $A$, we are free to use, in particular any set that suits us. So let's use $A = \{a\}$, a set with just one element.

Specifying a function $g_1: \{a\} \to B$ requires just picking out which element of $B$ we want to assign $a$ to, say:

$g_1(a) = b_1$.

Similarly, we specify $g_2$ by assigning $g_2(a) = b_2$.

Now suppose we have $f \circ g_1 = f\circ g_2$. Then:

$(f\circ g_1)(a) = (f\circ g_2)(a)$ (since equal functions must take equal values on the same domain element)
$f(g_1(a)) = f(g_2(a))$
$f(b_1) = f(b_2)$.

By Definition 2, we have that $f \circ g_1 = f \circ g_2 \implies g_1 = g_2$, so:

$g_1(a) = g_2(a)$
$b_1 = b_2$, and thus $f(b_1) = f(b_2) \implies b_1 = b_2$, and $f$ is injective. This is actually the easy part.

Definition 1 $\implies$ Definition 2:

Suppose $f(b_1) = f(b_2) \implies b _1 = b_2$. We need to show that for ANY set $A$ and ANY two functions $g_1,g_2: A \to B$ such that $f \circ g_1 = f \circ g_2$, we have $g_1 = g_2$. We will do this by way of contradiction.

Suppose, contrary to assumption, we have $f \circ g_1 = f \circ g_2$, but we do NOT have $g_1 = g_2$. Since these functions are not equal, there must be some $a \in A$ such that $g_1(a) \neq g_2(a)$.

However, since $f \circ g_1 = f \circ g_2$, we have:

$(f \circ g_1)(a) = (f \circ g_2)(a)$
$f(g_1(a)) = f(g_2(a))$

Taking $b_1 = g_1(a)$, and $b_2 = g_2(a)$ (since these are two elements of $B$, since $g_1,g_2: A \to B$), we have (by Definition 1), that $g_1(a) = b_1 = b_2 = g_2(a)$, a contradiction (by our choice of $a$). This completes the proof.

So the two definitions really are equivalent, and we can use whichever one we like. As you might have conjectured, we will prefer the latter, for reasons that hopefully will be clear later.

A small theorem:

Theorem 1: If $f:B \to C$ and $g: A \to B$ are functions such that $f \circ g$ is injective, then $g$ is injective.

Proof: Suppose $h_1,h_2: X \to A$ are two functions such that $g \circ h_1 = g \circ h_2$. It follows that:

$(f \circ g) \circ h_1 = f \circ (g \circ h_1) = f\circ (g \circ h_2) = (f \circ g) \circ h_2$.

Since $f \circ g$ is injective, we have $h_1 = h_2$, QED.

(Note: we have used the associativity of composition of functions without proof. The interested reader is encouraged to demonstrate this to their own satisfaction).

We will be more interested in the "mirror images" of these statements in our next post.

Questions and comments should be posted here:

http://mathhelpboards.com/commentary-threads-53/commentary-quot-de-mystifying-universal-mapping-properties-example-quotient-groups-quot-18583.html

 

[Deveno]

This thread is for comments/questions pertaining to:

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[Deveno]

In abstract algebra things often "come in pairs". The properties of injective functions we have just examined have a twin sister, which we turn our attention to now:

A function $f:B \to C$ is called onto or surjective, if every element in $C$ can be "traced" back to an origin in $B$ under the function $f$, that is:

For every $c \in C$ there is some $b \in B$ with $f(b) = c$. One can think of this as $f$ "covering" $C$. As promised, we have a similar "Definition 2" that goes along with this:

A function $f: B \to C$ is surjective $\iff$ for every pair of functions $k_1,k_2: C \to D$ such that $k_1 \circ f = k_2 \circ f$ we have $k_1 = k_2$.

Definition 1 $\implies$ Definition 2:

Suppose we have two functions $k_1,k_2: C \to D$ as above, with $k_1 \circ f = k_2 \circ f$, and $f$ is surjective. Let $c$ be ANY element of $C$. We will show that $k_1(c) = k_2(c)$, establishing that $k_1 = k_2$.

Since $f$ is surjective, we have that $c = f(b)$, for some $b \in B$. Since $k_1 \circ f = k_2 \circ f$, we have, in particular:

$(k_1 \circ f)(b) = (k_2 \circ f)(b)$
$k_1(f(b)) = k_2(f(b))$
$k_1(c) = k_2(c)$ (Ta da!) Again, this is sort of the "easy part".

Definition 2 $\implies$ Definition 1:

Suppose that for any set $D$ and any two functions $k_1,k_2:C \to D$ we have $k_1 \circ f = k_2 \circ f \implies k_1 = k_2$. Let us use these two particular functions, for the set $D = \{d_1,d_2\}$:

$k_1: C \to \{d_1,d_2\}$ given by $k_1(c) = d_1$ for all $c \in C$
$k_2: C \to \{d_1,d_2\}$ given by $k_2(c) = d_1$, if $c = f(b)$ for some $b \in B$, and $k_2(c) = d_2$, otherwise.

Then $(k_1 \circ f)(b) = k_1(f(b)) = d_1$, for all $b \in B$, and: $(k_2\circ f)(b) = k_2(f(b)) = d_1$ (by our definition of $k_2)$.

Thus, for all $b \in B$, $k_1 \circ f = k_2 \circ f$, and by Definition 2, $k_1 = k_2$. Thus, $f$ is surjective (re-read this until it is perfectly clear in your mind).

As before, we have a sister theorem to Theorem 1, which I will call (most imaginatively):

Theorem 2: If $f: B \to C$ and $k: C \to D$ are functions such that $k \circ f$ is surjective, then $k$ is surjective.

Proof: Let $h_1,h_2: D \to Y$ be two functions such that $h_1 \circ k = h_2 \circ k$. We seek to show that $h_1 = h_2$.

Now $h_1 \circ (k \circ f) = (h_1 \circ k) \circ f = (h_2 \circ k) \circ f = h_2 \circ (k \circ f)$. Since $k \circ f$ is surjective, we have $h_1 = h_2$, QED.

Theorem 2 may not seem like "a big deal" on the face of things, but we will use it like a weapon in the task ahead. We will be interested in proving things about certain functions without having to "delve into their inner workings." Be patient. In what follows we will be mostly concerned about surjective functions, but injective ones will occasionally pop by and say hello.

 

[Deveno]

So, as promised, we now turn our attention to quotient groups. I will give an *unusual* definition of them, and after playing with it for a while, we will finally see it is the same thing as the "regular way" of talking about them.

Why the long introduction on functions? Because instead of looking at a group *operation* (which is usually the STARTING point for talking about some kind of group), we are going to be looking at group HOMOMORPHISMS, instead. This focus on homomorphisms will allow us to prove some powerful theorems "without getting our hands dirty", so to speak. That is, instead of looking at what quotient groups ARE, we will focus on how they BEHAVE.

It will be assumed that the reader has had SOME exposure to the concept of a group, which is:

A set, usually called $G$.
A binary operation $\ast: G \times G \to G$ (usually written $(a,b) \mapsto a\ast b$), and called (for lack of a better term) "multiplication" that is:

1. Associative
2. Possesses an identity element (usually written $e$), for which $a \ast e = e \ast a = a$, for all $a \in G$
3. Possesses an inversion map $G \to G$ (usually denoted $\ ^{-1}$); that is, for every $a \in G$ there exists a unique $b \in G$ with $a \ast b = b \ast a = e$.

Given two groups $G,G'$, a homomorphism is a mapping $\phi: G \to G'$ such that $\phi(a \ast b) = \phi(a) \ast' \phi(b)$. Such a map is said to preserve products.

A subset $H$ of $G$ is said to be a subgroup of $G$ if $\ast$ restricted to $H$ is also a binary operation on $H$ (a condition called closure) that results in $H$ being a group in its own right.

One homomorphism that is particularly important is the homomorphism $G \to G'$ that sends all of $G$ to the identity $e'$ of $G'$. We will call this homomorphism the zero homomorphism, and denote it by $0$ when this will not cause confusion.

If a homomorphism $\phi: G \to G'$ is the 0-homomorphism when restricted to a subgroup $H$, will will say $\phi$ kills $H$.

So, here is our definition of (a) quotient group:

A quotient group $Q$ of a group $G$ by a subgroup $H$ is:

1. A group $Q$ together with a surjective homomorphism $\pi: G \to Q$ that kills $H$.
2. $\pi$ is universal among homomorphisms that kill $H$.

What we mean by 2 is the following:

If $\phi: G \to X$ is a homomorphism to ANY group $X$ such that $\phi$ kills $H$, then there exists a UNIQUE homomorphism $\psi: Q \to X$ such that $\phi$ "factors through $\pi$", that is:

$\phi = \psi \circ \pi$

We shall see that this actually imposes some special conditions on what "kind" of subgroup $H$ can be, because of the condition that $\pi$ be a homomorphism (which are SPECIAL functions), so that not every subgroup $H$ will be suitable for formng a quotient.

The more practical-minded among you, might think at this point-why would we do this? Well, often we might start with a "known" homomorphism, but we're only interested in "certain parts" of our group. The parts we're not interested in, we might "mod away", and it would be handy to know that this process induces a well-defined homomorphism on the resulting quotient. It turns out it does, and we will be able to apply our results to create such "induced homomorphisms" quickly, and painlessly. Stay tuned.

 

[Math Amateur]

Re: Commentary for "De-mystifying universal mapping properties: an example-quotient groups."

This thread is for comments/questions pertaining to:

http://mathhelpboards.com/math-notes-49/de-mystifying-universal-mapping-properties-example-quotient-groups-18582.html

Hi Deveno ... thank you for posting on this topic ... so helpful as the topic is certainly mystifying and highly unsettling ...

Indeed ... you write ...

"Often, in the study of algebraic objects certain things (like tensor products) are often defined primarily in terms of an universal mapping property. When one is used to "concrete objects" one can calculate with, this often comes as a shock to the system. One feels as if one is spinning something out of nothing, and has the feeling of "waiting for the real 'stuff' to show up." ... ... "

Exactly! You are right on the money ... ... precisely how I feel when approaching this material ... ... but on the other hand ... it is quite fascinating ... as the spider's web is to the fly ... ...

Now ... back to studying your posts ... ... Peter

 

[Math Amateur]

Re: Commentary for "De-mystifying universal mapping properties: an example-quotient groups."

This thread is for comments/questions pertaining to:

http://mathhelpboards.com/math-notes-49/de-mystifying-universal-mapping-properties-example-quotient-groups-18582.html

You write:https://www.physicsforums.com/attachments/5587
In the above definition A can be any set ... so don't we have to show that Definition 1 implies Definition 2 for any ... and hence all sets ...?

Can you clarify this matter ... ?

Peter

 

[Deveno]

Re: Commentary for "De-mystifying universal mapping properties: an example-quotient groups."

You write:
In the above definition A can be any set ... so don't we have to show that Definition 1 implies Definition 2 for any ... and hence all sets ...?

Can you clarify this matter ... ?

Peter

That would be the second half of the proof. I am proving Definition 2 implies Definition 1, here. Since Definition 2 is quite general, we lose nothing by using some particular instance of it.

 

[Math Amateur]

Re: Commentary for "De-mystifying universal mapping properties: an example-quotient groups."

That would be the second half of the proof. I am proving Definition 2 implies Definition 1, here. Since Definition 2 is quite general, we lose nothing by using some particular instance of it.

Oh ... OK ... yes, of course ... thanks for pointing that out ...

Peter

 

[Deveno]

So, let's forge ahead.

Our very first result (and one that is proved in essentially the same way for any object defined by a universal mapping property) is:

Any two quotients of a group $G$ by a subgroup $H$ are isomorphic.

In terms of how we have defined them, this means that we have two surjective homomorphisms:

$\pi_1: G \to Q_1$
$\pi_2: G \to Q_2$

that both kill $H$, and if we have any homomorphism $\phi:G \to K$ that kills $H$, there are unique homomorphisms:

$\psi_1:Q_1 \to K$
$\psi_2:Q_2 \to K$

such that $\phi = \psi_1 \circ \pi_1$ and $\phi = \psi_2 \circ \pi_2$.

So, for example, we can take $K = Q_2$ and $\phi = \pi_2$, and we thus have a unique homomorphism:

$\psi_1: Q_1 \to Q_2$ with $\psi_1 \circ \pi_1 = \pi_2$.

On the other hand, we can also take $K = Q_1$ and $\phi = \pi_1$, to establish we have a unique homomorphism:

$\psi_2: Q_2 \to Q_1$ with $\psi_2 \circ \pi_2 = \pi_1$.

Thus $(\psi_1 \circ \psi_2) \circ \pi_2 = \psi_1 \circ (\psi_2 \circ \pi_2) = \psi_1 \circ \pi_1 = \pi_2 = 1_{Q_2} \circ \pi_2$.

Since $\pi_2$ is surjective, by Definition 2 we have: $\psi_1 \circ \psi_2 = 1_{Q_2}$, and a similar argument shows:

$\psi_2 \circ \psi_1 = 1_{Q_1}$.

Since everything in sight is a homomorphism, all our compositions are, (note to the interested reader: *prove* the composition of two homomorphisms is again a homomorphism-it's not hard) and both $1_{Q_1},1_{Q_2}$ are surjective, it follows that $\psi_1$ and $\psi_2$ are surjective by Theorem 2, and it follows that they are injective by Theorem 1.

Thus $\psi_1: Q_1 \to Q_2$ is an isomorphism of $Q_1$ with $Q_2$ (and, of course, $\psi_2$ is an isomorphism of $Q_2$ with $Q_1$). Note we have proved this without knowing a SINGLE thing about ANY of the elements (except that groups have an identity) of $G$ OR $H$, just using some basic facts about functions.

But let's talk about the mapping property itself-what does it MEAN for a homomorphism $\phi: G \to K$ to "factor through" $\pi:G \to Q$?

To answer this, it helps to look at ordinary functions. The answer may seem surprising: a surjective function is an "equivalence relation" (or partition of a set) in disguise.

For what does a function "do"? It takes domain elements and "sends them somewhere", and we can "sort" the domain elements according to "where they get sent". For example, the function:

$f: \Bbb R \to \Bbb R_0^+$ with $f(x) = |x|$, takes negative and positive real numbers with the same magnitude, and "lumps them together". We can define an equivalence relation on $\Bbb R$ by:

$x \sim y$ if $|x| = |y|$.

For example, the equivalence class (partition subset) corresponding to $4$ is $\{-4,4\}$. This is by no means limited to "just" some special kind of surjective function, we can for ANY function:

$f:A \to f(A)$ define an equivalence relation $\sim_f$ by:

$a_1 \sim_f a_2 \iff f(a_1) = f(a_2)$. I leave it to the reader to establish this is indeed a bona-fide equivalence relation.

So note that if we have:

$h: A \to C$ such that there exists $g:f(A) \to C$, with $h = g \circ f$, it must be the case that for any $a \in A$, if we take the equivalence class of $a$ under $\sim_f$ (let's call this $[a]_f$), $h$ has to be CONSTANT on $[a]_f$.

Because if $a' \in [a]_f$ (that is, $f(a') = f(a)$), then:

$h(a') = (g\circ f)(a') = g(f(a')) = g(f(a)) = h(a)$.

We can re-phrase this as: $h$ RESPECTS the partition (of equivalence classes) of $A$ induced by $f$.

So, if $h$ itself is ALSO surjective, it must represent a MERGING (perhaps a trivial one, where all equivalence classes get lumped together, or one where they are left unchanged) of equivalence classes of $A$. Another way of saying this is that $f$ induces a REFINEMENT of the partition induced by $h$. This is what it means for $h$ to "factor through $f$". In effect, $h$ has "blurrier vision" than $f$, and can only distinguish elements of $A$ "as well as $f$ can" (or maybe worse).

If we denote the set of equivalence classes of $A$ induced by $f$ as $A/f$, we have the following important result:

there is a bijection $A/f \to f(A)$ (we can think of a bijection as a "set-isomorphism"). The bijection is:

$[a]_f \mapsto f(a)$.

Note that $f:A \to f(A)$ respects its own partition (by definition), and that factors through the map $a \mapsto [a]_f$ -lets's call this map $q$-and the map $[a]_f \mapsto f(a)$, say $k$. Since $f = k \circ q$ it follows (from Theorem 2 and the surjectivity of $f$) that $k$ is surjective. To see that $k$ is injective, suppose $k([a_1]_f) =(k \circ q)(a_1) = (k \circ q)(a_2) = k([a_2]_f)$.

Then $f(a_1) = f(a_2)$, and thus $[a_1]_f = [a_2]_f$.

(Important sidebar: $k$ is defined on $A/f$, not $A$. So for $k$ to be "well-defined" as a function, it has to be "constant on each equivalence class of $\sim_f$". Fortunately, this is indeed the case-if $a' \in [a]_f$, then $f(a') = f(a)$, and $k([a']_f) = f(a') = f(a) = k([a])_f$, that is, $k$ only depends on $[a]_f$, and not which $a' \in [a]_f$ we use).

So we can regard $f: A \to f(A)$ as establishing a "set-isomorphism" between $f(A)$ and the partition $A/f$.

Now let's revisit this in terms of group homomorphisms. We can now regard:

$\pi: G \to Q$

as a mapping $g \to [g]_{\pi}$ (more accurately, as establishing a group isomorphism between $G/\pi$ and $Q$), and the homomorphism property says that:

$[gh]_{\pi} = [g]_{\pi}[h]_{\pi}$.

An equivalence relation which satisfies this type of respecting of an operation is called a CONGRUENCE. It says that we can "multiply" and "clump together" in EITHER ORDER, and we arrive in the same place. We shall look at congruences in greater detail in the next post.

 

[Math Amateur]

Re: Commentary for "De-mystifying universal mapping properties: an example-quotient groups."

This thread is for comments/questions pertaining to:

http://mathhelpboards.com/math-notes-49/de-mystifying-universal-mapping-properties-example-quotient-groups-18582.html

This series of posts continues to be just amazingly helpful ...

These posts cover what is to me an unfamiliar and difficult area with a clarity that seems to me to be completely lacking in the standard texts that seem to assume you already have a basic understanding of UMPs ...

So ... THANK YOU ...

Peter

 

[Deveno]

So, let's return to an examination of what happens with a surjective homomorphism and how congruences play into this.

Fact 1: Every surjective homomorphism induces a congruence, and vice-versa.

Before we begin, let me introduce the *usual* definition of a congruence on a group.

A congruence on a group $G$ is an equivalence relation $\sim$ such that whenever:

$a\sim a'$ and $b \sim b'$, we have $ab \sim a'b'$ (for any $a,a',b,b' \in G$).

We shall refer to the equivalence class of an element $g \in G$ as its congruence class, and denote it so $[g]_{\sim}$ or simply $[g]$ when the equivalence relation is understood.

Now if we have a surjective homomorphism (like our $\pi: G \to Q$), we can form the equivalence relation induced by $\pi$. We shall now show this is, in fact, a congruence (something stated without proof in the prior post).

So suppose $a \sim_{\pi} a'$ and $b \sim_{\pi} b'$. What this means is $\pi(a) = \pi(a')$ and $\pi(b) = \pi(b')$.

Then $\pi(ab) = \pi(a)\pi(b) = \pi(a')\pi(b') = \pi(a'b')$, so $ab \sim_{\pi} a'b'$ (see how homomorphisms make this easy?).

On the other hand, suppose we have a congruence, $\sim$. We need to somehow come up with some other group, say $H$, and display a surjective homomorphism $\phi: G \to H$.

Well, the natural candidate for a surjective mapping is $\phi: G \to G/\sim$ (the set of congruence classes of $G$) given by $\phi(g) = [g]_{\sim}$, but we need to fashion $G/\sim$ into a group.

If we try:

$[a]\ast = [ab]$, we have that this is well-defined on $G/\sim$ because $\sim$ is a CONGRUENCE. So let us see if it obeys all 3 group axioms:

1. $\ast$ is associative:

$([a]\ast)\ast [c] = [ab]\ast [c] = [(ab)c] = [a(bc)] = [a]\ast[bc] = [a]\ast(\ast[c])$.

2. $[e]$ (where $e \in G$ is $G$'s group identity) is an identity for $G/\sim$:

$[a]\ast[e] = [ae] = [a]$
$[e] \ast [a] = [ea] = [a]$

3. $[a^{-1}]$ is an inverse for $[a]$:

$[a]\ast[a^{-1}] = [aa^{-1}] = [e]$
$[a^{-1}]\ast[a] = [a^{-1}a] = [e]$

So now we have a surjective mapping $\phi:G \to G/\sim$, and all we must do is verify that $\phi$ is a homomorphism. But this is immediate from our definition of $\ast$:

$\phi(ab) = [ab] = [a]\ast = \phi(a)\ast\phi(b)$.

These two processes functions as sorts of "inverses" to each other:

Suppose we start with a surjective homomorphism $\pi: G \to Q$. We can form the equivalence relation induced by $\pi$, and consider $G/\pi$.

We can define a product on $G/\pi$ like so:

$(\pi^{-1}(\pi(a))\ast (\pi^{-1}(\pi(b))) = \pi^{-1}(\pi(ab))$, since:

$[a]_{\pi} = \pi^{-1}(\pi(a))$, by definition, this just says: $[a]_{\pi}\ast_{\pi} = [ab]_{\pi}$.

Thus this multiplication is just the congruence class multiplication we used to show $G/\sim$ is a group, and we know $\sim_{\pi}$ is a congruence by our discussion above.

To go further we will need one more piece of the puzzle:

Fact 2: If $\sim$ is a congruence on a group $G$, then the congruence class of the identity $[e]_{\sim}$, is a subgroup of $G$.

This is actually sort of surprising, since nothing we discussed about congruences would lead us to conjecture this fact. However, since the identity of $G$ is a "special element", we might suppose that similarly $[e]_{\sim}$ is *also* somehow "special".

To prove fact 2, we have to first show that if $a \in [e]_{\sim}$ and $b \in [e]_\sim$ that $ab \in [e]_{\sim}$.

Note that since $\sim$ is a CONGRUENCE, and we have $a \sim e$ AND $b \sim e$, that we have:

$ab \sim ee = e$, so $ab \in [e]_{\sim}$. Well, that was easy.

Of course, it is trivial that $e \in [e]_{\sim}$ (because $\sim$ is an equivalence, and these are reflexive), so now we seek to prove that if $a \in [e]_{\sim}$ that $a^{-1} \in [e]_{\sim}$ also.

Now $a \sim e$, and certainly $a^{-1} \sim a^{-1}$, thus, since we have a congruence:

$e = aa^{-1} \sim ea^{-1} = a^{-1}$, thus $a^{-1} \in [e]_{\sim}$.

Thus $[e]_{\sim}$ is a subgroup of $G$.

Now, by design, we have that $\pi:G \to Q$ is a surjective homomorphism that kills the subgroup $[e]_{\pi} = \pi^{-1}(\pi(e))$. Our claim is now:

$(Q,\pi)$ is a quotient of $G$ by $[e]_{\pi}$.

For suppose $\phi:G \to K$ is a surjective homomorphism that kills $[e]_{\pi}$. Define $\psi: Q \to K$ by:

$\psi(q) = \phi(a)$ where $\pi(a) = q$.

For this to be well-defined, we need to show $\phi$ is constant on $[a]_{\pi}$, because we might have many such potential $a \in G$.

So suppose $\pi(a) = \pi(b) = q$.

Then $\pi(a)(\pi(b))^{-1} = qq^{-1} = e_Q$. Since $\pi$ is a homomorphism, we have:

$\pi(b)\pi(b^{-1}) = \pi(bb^{-1}) = \pi(e_G) = e_Q$, so by uniqueness of inverses $\pi(b^{-1}) = \pi(b)^{-1}$.

So $e_Q = \pi(a)\pi(b)^{-1} = \pi(a)\pi(b^{-1}) = \pi(ab^{-1})$.

Hence $ab^{-1} \in [e]_{\pi}$. Since $\phi$ kills $[e]_{\pi}$, and is *also* a homomorphism, we have:

$\phi(ab^{-1}) = \phi(a)\phi(b^{-1}) = \phi(a)\phi(b)^{-1} = e_K$.

Hence $\phi(a) = \phi(b)$, so our $\psi$ is well-defined (make sure you follow this!).

Thus $\phi(a) = \psi(q) = \psi(\pi(a))$, that is $\phi$ factors through $\pi$, and it is clear the way we have defined $\psi$ that this is the ONLY mapping that will work.

However, we still need to show that $\psi$ is indeed a homomorphism:

So choose $a,a' \in G$ such that $\pi(a) = q, \pi(a') = q'$ (we can do this PRECISELY because $\pi$ is surjective).

Then $\psi(qq') = \psi(\pi(a)\pi(a')) = \psi(\pi(aa')) = \phi(aa') = \phi(a)\phi(a') = \psi(\pi(a))\psi(\pi(a')) = \psi(q)\psi(q')$.

This establishes that $\pi: G \to Q$ is a quotient of $G$ by $[e]_{\pi}$.

Now BY THE UNIVERSAL PROPERTY OF THE QUOTIENT, since the mapping $G \to G/\pi$ is ALSO a surjective homomorphism that kills $[e]_{\pi}$ we conclude THERE EXISTS a (unique!) homomorphism $G/\pi \to Q$. Since we saw in the last post that the mapping:

$A/f \to f(A)$ given by $[a]_f \mapsto f(a)$ is a bijection, this homomorphism is thus an isomorphism, that is:

$G/\pi \cong Q$.

This is the fundamental isomorphism theorem for groups, and the subgroup $[e]_{\pi}$ is usually called the *kernel* of $\pi$.

In short, when we form the congruence induced by a surjective homomorphism $\pi:G \to Q$, and then form the surjective homomorphism induced by the congruence $G/\pi$, we don't "exactly" recover our original homomorphism, but we get something almost as good, an isomorphism. This is an example of what is called in category theory, a "natural equivalence".

We will see that subgroups that induce a congruence like this are *special*, possessing qualities not all subgroups do.