Let f : X → Y where X ⊂ R^{n}, Y ⊂ R^{m} Graph of f = {(x, f(x)) | x ∈ X} ⊂ R^{n} × R^{m} Domain of f = X, Codomain of f = Y, Image of f = f(X). f is a function if for all x in domain of f, f(x) has a unique value.

I.e, for all x, y ∈ X, if x = y, then f(x) = f(y) f is 1:1 if f(x) = f(y) implies x = y.

f gives a one-to-one correspondence between X and f(X). Given b ∈ Y, f(x) = b has at most one solution

Side-note: f(x) = b has exactly one solution if b ∈ f(X). . Side-note: f(x) = b has no solution if b ∉ f(X). f is onto if f(X) = Y (i.e., image of f = codomain of f).

Given b ∈ Y, f(x) = b has at least one solution. Ex 1: f : R^{n} → R, f(x) = ||x||

Domain = R^{n} Codomain = R Image = [0, ∞) f is not 1:1: f(1, 0, ..., 0) = 1 = f(0, 1, 0, ..., 0) f is not onto: f(x) = −1 has no solution.

or codomain of f = R ≠ [0, ∞) = image of f Ex 2: g(x, y) = (x^{2}y, x^{4} − y, x^{6})

Domain = R^{2} Codomain = R^{3} g is not 1:1: g(1, 0) = (0, 1, 1) = g(−1, 0) g is not onto: g(x, y) = (0, 1, −1) has no solution.

Ex 3: h(x) = ( \begin{matrix}1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} ) ( \begin{matrix}x \\ y \\ z \end{matrix} )

I.e, h(x) = (x + 2y + 3z, 4x + 5y + 6z).

Domain = Codomain = Image = Is h onto? Is h 1:1?

How many solutions does h(x) = b have?

I.e., how many solutions does ( \begin{matrix}1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} ) ( \begin{matrix}x \\ y \\ z \end{matrix} ) = ( \begin{matrix}b_{1} \\ b_{2} \end{matrix} ) have?

I.e, how many solutions does the following system of equations have:

x + 2y + 3z = b_{1},
4x + 5y + 6z = b_{2}.

I.e., does ( \begin{matrix}1 \\ 4 \end{matrix} ) x + ( \begin{matrix}2 \\ 5 \end{matrix} ) y + ( \begin{matrix}3 \\ 6 \end{matrix} ) z span all of R^{2}?