Surjection(surjectivity): everything in the codomain gets hit by something
Definition Let A and B be sets and f: A \rightarrow B be a function. f is surjective (or onto) iff Im_f(A) = B.
(\forall b \in B)(\exists a \in A)(f(a) = b)
f:A \rightarrowtail B
then |A|>=|B| (B got all mapped even though there may be a overlap)
Definition: let A, and B be set and f: A \to B be a function, we say that f is injective(1-to-1).
(\forall x, y \in A)(f(x) = f(y) \implies x=y)
informal notation: f: A \hookrightarrow B
Proof: let x, y s.t. f(x)=f(y), show x=y. You can show f(x)!=f(y) to by pass case check for piece-wise function.
then |A|<=|B| (one to one, but not all B gets mapped)
Definition: let A, and B be set and f: A \to B be a function, we say that f is bijection iff f is both injection and surjection.
Definition: Let A, B, C be sets and f:A\to B \land g:B\to C be functions. The function k:A\to C are defined by (\forall a \in A)(h(a) = g(f(a))) is called the composition of g and f, denoted h=g \circ f.
Theorem: Let A, B, C, D be sets and f:A\to B \land g:B\to C, h:C\to D be functions. Then f \circ (g \circ f) = (h \circ g) \circ f
observe: if f: A\to A, then id_A \circ f = f \circ id_A = f
id_A: A \to A, a |-> a TODO what is this notation TODO what is identity on a function
Definition: Let A, B be sets and f: A \to B and g: B \to A be functions. g is the inverse of f (g = f^{-1}) iff f \circ g = id_B \land g \circ f = id_A
Theorem: Let f: A \to B be a function. f is invertible iff f is a bijection.
prove forward: f is invertible -> f is a bijection
prove backward: assume 1-to-1, onto.
Corollary: if f is invertible, then f^-1 is unique
Claim: f: \mathbb{R} / \{3\} \to \mathbb{R} / \{1\} is a bijection by f(x) = \frac{x-2}{x-3}
Scratch:
solve x = \frac{3y - 2}{y - 1}
so x \neq 3 \land y \neq 1
Proof:
define g(x) = \frac{3x - 2}{x - 1}
observe (\forall x \in \mathbb{R} \ \{1\})(g(x) \in \mathbb{R}) because x \neq 1
observe g(x) \neq 3 because g(x)=3 \iff \frac{3x-2}{x-1} = 3 \iff 3x-2=3x-3 \iff -2=-3
so g is well-defined
then g(f(x)) = \frac{3\frac{3-2}{x-3}-2}{\frac{x-2}{x-3}-1} = x so g \circ f = id_{\mathbb{R} \ \{3\}} holds
show the same thing for f(g(x))
f is invertible -> f is a bijection
Table of Content