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- Feb 15, 2012

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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:

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

__Theorem 1____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:

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