Lecture 4: proofs and functions

reading: MCS 4.3-4.4
define 'jectivity (injectivity, bijectivity, surjectivity)
define composition, inverses.
discuss connection between 'jectivity and inverses

'jectivity

Definitions: let $f : A → B$ . Then $f$ is

injective (or "one to one") if for all $x_1$ , $x_2 \in A$ , if $f(x_1) = f(x_2)$ then $x_1 = x_2$
surective (or "onto") if for all $y \in B$ , there exists an $x \in A$ such that $f(x) = y$
bijective if it is both injective and surjective

composition

Definition: if $f : A → B$ and $g : B → C$ then the composition of g with f (written $g \circ f$ ) is the function $(g \circ f) : A → C$ is given by $(g \circ f)(x) = g(f(x))$ .

Note that $g \circ f$ is different from $f \circ g$ .

inverses

Definition: The identity function on X is the function $id : X → X$ given by $id(x) = x$ .

Definition: If $f : A → B$ then $g$ is a left inverse of $f$ if $g \circ f = id$ . In other words, if for all $x \in A$ , $g(f(x)) = x$ (Note: $g$ is a left inverse because you write it on the left).

Similarly, $g$ is a right inverse if $f \circ g = id$ .

'jectivity and inverses

Claim: $f$ has a right inverse if and only if it is surjective.

Proof: We must prove that if $f$ has a right inverse, then it is surjective, and also that if $f$ is surjective, then it has a right inverse.

(RI ⇒ surj) Suppose that $f$ has a right inverse $g$ . We will show that $f$ is surjective by contradiction. Suppose that $f$ is not surjective. Then there exists some $y$ such that for all $x$ , $f(x) \neq y$ (). Now, since $g$ is a right inverse of $x$ , $f(g(y)) = y$ . But this contradicts (), so our original assumption (that $f$ is not surjective) must be false.

(surj ⇒ RI) Suppose $f$ is surjective. We will show that $f$ has a right inverse by constructing it. Let $g : B → A$ be defined as follows. Given any $y \in B$ we know that there exists some $x \in A$ such that $f(x) = y$ (since $f$ is surjective). Choose one of them, and define $g(y)$ to be $x$ .

I claim $g$ is a right inverse of $f$ . Indeed, for any $y \in B$ , we have $f(g(y)) = y$ by the definition of $g$ .

This concludes the proof.

Note: it turns out that $f$ is injective if and only if it has a left inverse, and $f$ is bijective if and only if it has a two-sided inverse. This will be on the next homework.