If $\alpha:S\rightarrow T$ and $\beta:T\rightarrow U$ then the composition (or composite) of $\alpha$ and $\beta$ is \( \beta\circ\alpha:S\rightarrow U \) and \( (\beta\circ\alpha)(x)=\beta(\alpha(x)). \) Note carefully, in $\beta\circ\alpha$, it is $\alpha$ the mapping on the right, that is applied first.

Let $S=\{x,y,z\},$ $T=\{1,2,3\},$ $U=\{a,b,c\},$ $\alpha:(x,y,z)\mapsto(2,1,3),$ $\beta:(1,2,3)\mapsto(b,c,a).$ Then $\beta\circ\alpha:(x,y,z)\mapsto(c,b,a)$

It is possible for $\beta\circ\alpha\ne\alpha\circ\beta.$ Composition is not necessarily commutative. Let $\alpha:\R\rightarrow \R,$ $\beta:\R\rightarrow \R,$ $\alpha(x)=x^2+2,$ $\beta(x)=x-1.$ Then $(\alpha\circ\beta)(0)=3$ but $(\beta\circ\alpha)(0)=1.$

Assume that $\alpha:S\rightarrow T$ and $\beta:T\rightarrow U.$
  1. If $\alpha$ and $\beta$ are onto, then $\beta\circ\alpha$ is onto.
  2. If $\beta\circ\alpha$ is onto, then $\beta$ is onto.
    $\alpha$ is not onto iff $\beta$ is not one-to-one.
  3. If $\alpha$ and $\beta$ are one-to-one, then $\beta\circ\alpha$ is one-to-one.
  4. If $\beta\circ\alpha$ is one-to-one, then $\alpha$ is one-to-one.
    $\beta$ is not one-to-one iff $\alpha$ is not onto.

A mapping $\beta:T\rightarrow S$ is an inverse of $\alpha:S\rightarrow T$ if both $\beta\circ\alpha=\iota_S$ and $\alpha\circ\beta=\iota_T.$ A mapping is invertible if it has an inverse.

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

The mapping $\alpha$ in is invertible. Its inverse is the mapping $\gamma$ defined by $\gamma(1)=y,$ $\gamma(2)=x,$ $\gamma(3)=z.$ For example, $(\gamma\circ\alpha)(x)$ $=\gamma(2)=x$ $=\iota_S(x),$ and $(\alpha\circ\gamma)(1)$ $=\alpha(y)=1$ $=\iota_T(1).$ There are also four other equations to be checked, those involving $(\gamma\circ\alpha)(y),$ $(\gamma\circ\alpha)(z),$ $(\alpha\circ\gamma)(2),$ $(\alpha\circ\gamma)(3).$ In terms of the diagram in , $\gamma$ is obtained by reversing the direction of the arrows under $\alpha.$

The mapping $\alpha:\R\rightarrow\R$ defined by $\alpha(x)=x^2$ is not invertible. It fails on both counts of the following theorem.

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

Some authors write mappings on the right rather than on the left. Our $\alpha(x)$ becomes, for them, $x\alpha.$ Then in $\beta\circ\alpha$ it is $\beta,$ the mapping on the left, that is applied first, because $x(\beta\circ\alpha)=(x\beta)\alpha.$