Math and science::Algebra::Aluffi

Injectivity and surjectivity

Injective

A function $f : A \to B$ is injective iff

[

(\forall a^{'} \in A) (\forall a^{″} \in A) ?

]

Injections are often denoted like [ $A ? B$ ].

Surjective

A function $f : A \to B$ is surjective iff

[

(\forall b \in B) (\exists a \in A) ?

]

Surjections are often denoted like [ $A ? B$ ].

Bijective

A function $f$ that is both injective and surjective is said to be bijective. A bijective function is called a bijection or a one-to-one correspondence or an isomorphism of sets; it is is often denoted like $f : A \tilde{\to} B$ .

If there is a bijection between sets $A$ and $B$ , then each element $a$ of $A$ can be matched with exactly one element $b$ of $B$ . $A$ and $B$ are said to be [...]. The statement that $A$ and $B$ have such a relationship is often denoted as $A ≅ B$ .

From the perspective of inverses

If $f : A \to B$ is a bijection, it can be 'flipped' to define a function $g : B \to A$ . This is a flipping of $Γ_{f}$ , the graph of $f$ , along it's diagonal. $f$ being a bijection insures that such a flipping action produces a valid function. The function $g$ has two interesting properties:

[

g \circ f = ?

]

[

f \circ g = ?

]

These properties correspond to the statement that the following two diagrams compute:

Bijections have inverses

The first identity above states that $g$ is a [...]-inverse of $f$ . The second identity states that $g$ is a [...]-inverse of $f$ . Combined we say that $g$ is the inverse of $f$ , and denote it as [...]. Thus, bijections have inverses.

If a function has an inverse, is it a bijection?

This is [true or false?]. We can, in fact, break this statement into two pieces.

Let $A$ and $B$ be sets, with $A \neq \emptyset$ , and let $f : A \to B$ be a function. Then the following bi-implications hold:

$f$ has a left-inverse if and only if it is [...].
$f$ has a right-inverse if and only if it is [...].

Can you think of the proof of these two bi-implications?

Futhermore, if $f$ has a left-inverse $g_{l}$ and right-inverse $g_{r}$ , then these inverses must be the same function! This can be shown by considering:

[

g_{l} = ? = g_{l} (f g_{r}) = (g_{l} f) g_{r} = ? = g_{r}

]

TODO: add diagram

18.09.2020