Definition (Page 15). If α:S→T and β:T→U then the composition (or composite) of α and β is β∘α:S→U and (β∘α)(x)=β(α(x)). Note carefully, in β∘α, it is α the mapping on the right, that is applied first.

Example 2.1. Let S={x,y,z}, T={1,2,3}, U={a,b,c}, α:(x,y,z)↦(2,1,3), β:(1,2,3)↦(b,c,a). Then β∘α:(x,y,z)↦(c,b,a)

Example 2.2. It is possible for β∘α≠α∘β. Composition is not necessarily commutative. Let α:R→R, β:R→R, α(x)=x2+2, β(x)=x−1. Then (α∘β)(0)=3 but (β∘α)(0)=1.

Theorem 2.1. Assume that α:S→T and β:T→U.
  1. If α and β are onto, then β∘α is onto.
  2. If β∘α is onto, then β is onto.
    Guevara Note 2.1. α is not onto iff β is not one-to-one.
  3. If α and β are one-to-one, then β∘α is one-to-one.
  4. If β∘α is one-to-one, then α is one-to-one.
    Guevara Note 2.2. β is not one-to-one iff α is not onto.

Definition (Page 17). A mapping β:T→S is an inverse of α:S→T if both β∘α=ιS and α∘β=ιT. A mapping is invertible if it has an inverse.

Theorem (Page 17). If a mapping is invertible, then its inverse is unique. (Problem 4.13).

Example 2.3. The mapping α in Example 2.1 is invertible. Its inverse is the mapping γ defined by γ(1)=y, γ(2)=x, γ(3)=z. For example, (γ∘α)(x) =γ(2)=x =ιS(x), and (α∘γ)(1) =α(y)=1 =ιT(1). There are also four other equations to be checked, those involving (γ∘α)(y), (γ∘α)(z), (α∘γ)(2), (α∘γ)(3). In terms of the diagram in Example 2.1, γ is obtained by reversing the direction of the arrows under α.

Example 2.4. The mapping α:R→R defined by α(x)=x2 is not invertible. It fails on both counts of the following theorem.

Theorem 2.2. A mapping is invertible iff it is both one-to-one and onto.

Warning (Page 18). Some authors write mappings on the right rather than on the left. Our α(x) becomes, for them, xα. Then in β∘α it is β, the mapping on the left, that is applied first, because x(β∘α)=(xβ)α.