Composition of Functions and Inverse Functions

Direct Knowlege Contributor David A. Smith
  • By David A. Smith, Founder & CEO, Direct Knowledge
  • David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

Composition

Theorem. Let $f:X\to Y$ be a function. If $g:Y\to Z$ and $g\circ f$ is injective, then $f$ is injective.

Proof. Let $x_1, x_2\in X$. Then $$ f(x_1)=f(x_2) \Longrightarrow (g\circ f)(x_1)=(g\circ f)(x_2) \Longrightarrow x_1=x_2 $$ shows that $f$ is injective.

Theorem. Let $f:X\to Y$ be a function.  If $g:Y\to Z$ and $g\circ f$ is surjective, then $g$ is surjective

Proof. Let $z\in Z$. Then there exists $x\in X$ such that $(g\circ f)(x)=z$. Thus it follows $g(f(x))=z$ shows, for all $z\in Z$ there exists $y$ (namely $y=f(x)$) such that $g(y)=z$ and so $g$ is surjective.

Theorem. Let $f:X\to Y$ be a function.  If $f$ is injective and $g:Y\to Z$ is injective, then $g\circ f$ is injective.

Proof. Let $x_1, x_2\in X$. Then $$ (g\circ f)(x_1)=(g\circ f)(x_2) \Longrightarrow f(x_1)=f(x_2)  \Longrightarrow x_1= x_2  $$ shows that $g\circ f$ is injective. 

Theorem. Let $f:X\to Y$ be a function.  If $f$ is surjective and $g:Y\to Z$ is surjective, then $g\circ f$ is surjective. 

Proof. Let $z\in Z$. Since $g$ is surjective there exists $y\in Y$ such that $g(y)=z$. Since $f$ is surjective there exists $x\in X$ such that $f(x)=y$. Thus $(g\circ f)(x)=z$ and so $g\circ f$ is surjective. 

Theorem. Let $f:X\to Y$ be a function.  Then $f$ is injective if and only if given any functions $g,h:Y\to X$, if $f\circ g=f\circ h$, then $g=h$.

Proof. Assume $f$ is injective. Let $g,h:Y\to X$ and assume $f\circ g= f\circ h$. Let $y\in Y$. Then $(f\circ g)(y)=(f\circ h)(y)$. Since $f$ is injective it follows $g(y)=h(y)$ for all $y\in Y$.  Conversely, assume $g,h:Y\to X$, $f\circ g=f\circ h \implies g=h$ holds. Let $x_1, x_2\in X$ and assume $x_1\neq x_2$. Assume, for a contradiction, $f(x_1)=f(x_2)$. Let $g:Y\to X$ be defined by $g(y)=x_1$ for all $y\in Y$. Let $h:Y\to X$ be defined by $h(y)=x_2$ for all $y\in Y$. Notice $$ (f\circ g)(y)=f(g(y))=f(x_1)=f(x_2)=(f\circ h)(y) $$ for all $y\in Y$. Thus, $f\circ g=f\circ h$, yet $g\neq h$ since $g(y)=x_1\neq x_2=h(y)$. Therefore, $f(x_1)\neq f(x_2)$ and so $f$ is injective.

Theorem. Let $f:X\to Y$ be a function.  Then $f$ is surjective if and only if given any functions $g,h:Y\to X$, if $g\circ f=h\circ f$, then $g=h.$

Proof. Assume $f$ is surjective and let $g:Y\to Z$ and $h:Y\to Z$ be functions such that $g\circ f=h\circ f$. Let $y\in Y$. Since $f$ is surjective, there exists $x\in X$ such that $y=f(x)$. Then $$g(y) =g(f(X)) =h(f(x)) =h(y)$$ as needed to show $g=h$. Conversely, and for contrapositive, suppose  $f$ is not surjective. Then there exists $y_1\in Y$ such that $y_1=f(x)$ does not hold for all $x\in X$. Let $Z=\{a,b\}$ and let $g$ and $h$ be defined by $g(y)=a$ for all $a\in Y$ and  $$h(y)=\begin{cases} a & \text{ if } y\neq y_1 \\ b & \text{ if } y=y_1. \end{cases} $$ Then we have $g\neq h$ such that $g\circ f=h\circ f$. Thus $f$ is not right cancelable.

Theorem. Let $f:X\to Y$ be a function.  Then $f$ is injective if and only if there exists a function $g:Y\to X$ with $g\circ f=I_X$ ($f$ has a left inverse).

Proof. Assume $X\neq \emptyset$ and also assume $f:X\to Y$ is injective.  Let $y\in Y$.  Either $y\in f(X)$ or $y\notin f(X)$.  Define $g:Y\to X$ as follows \begin{equation} g(y)= \begin{cases} x & y\in f(X) \text{ and } f(x)=y \\ x_0 & y \notin f(X)  \end{cases} \end{equation} where $x_0$ is a fixed element of $X\neq \emptyset$. Then $g$ is defined for all $y\in Y$. If $y’\in Y$ with $x_1=g(y’)\neq g(y’)=x_2$, then $y’=f(x_1)\neq f(x_2)=y’$, since $f$ is injective. Thus $f$ is injective implies $g$ is a function. If $x\in X$, then $$(g\circ f)(x)=g(f(x))=g(y)=x, \qquad \text{ for some $y\in f(X)$} $$ which shows $g\circ f=I_X$. Conversely, assume $f:X\to Y$ and there exists $g:Y\to X$ such that $g\circ f=I_X$. Let $x_1,x_2\in X$ and assume $f(x_1)=f(x_2)$. Then $$x_1=(g\circ f)(x_1)=g(f(x_1))=g(f(x_2))=(g\circ f)(x_2)=x_2 $$ which shows $f$ is injective.

Theorem. Let $f:X\to Y$ be a function.  Then $f$ is surjective if and only if there exists a function $h:Y\to X$ with $f\circ h=I_Y$ ($f$ has a right inverse). 

Proof. Assume there exist $h:Y\to X$ with $f \circ h=I_Y$. Let $b\in Y$. Then $$b=(f\circ h)(b)=f(h(b))=f(a)$$ where $a\in X$, shows $f$ is surjective.  Conversely, follows using the Axiom of Choice.  Suppose $f$ is surjective. Then $f^{-1}(b)\subseteq X$ is a nonempty set for every $b\in Y$. For each $b\in Y$ choose $a_b\in f^{-1}(b)$. Then the map $h:Y\to X$ defined by $h(b)=a_b$ is such that $f\circ h=I_Y$ since $(f\circ h)(y)=f(h(y))=f(a_y)=y$.

Inverse Functions

If the inverse relation $f^{-1}$ is also a function, then we say $f$ is an invertible function, or just invertible

Theorem. A function $f:X\to Y$ is invertible if and only if it is injective.  If $f$ is invertible then $f^{-1}$ is also invertible and $(f^{-1})^{-1}=f$.

Proof. Let $f$ be invertible, then $f^{-1}$ is a function. It follows that $f^{-1}(f(x))=x$ for all $x\in X$.  If $a=f^{-1}(y_1)$ and $a=f^{-1}(y_2)$ we have $y_1 = f(a)$ and $y_2 =f(a)$. So $f$ is injective. Let $f$ be injective. If $af^{-1}y_1$ and $af^{-1}y_2$ we have $y_1 f(a$ and $y_2 fa$. Therefore, $y_1=y_2$ and we have proven that $f^{-1}$ is a function.  We know that $(f^{-1})^{-1}=f$ by previous theorem and so $f^{-1}$ is also invertible.

Definition. Let $X$ and $Y$ be sets. A function $f$ is called bijective, if $f$ is both injective and surjective.

Theorem. Let $f:X\to Y$ be a function.  Then $f$ is bijective if and only if there exists a unique function $f’:Y\to X$ such that both $f’\circ f=I_X$ and $ f \circ f’ = I_Y$.

Proof. If $f$ is bijective, then $f$ is injective and surjective; and thus there exists functions $g$ and $h$ such that $g\circ f=I_X$ and $ f \circ h = I_Y$. Notice $$ g=g \circ I_Y=g \circ (f \circ h)=(g\circ f)\circ h=I_X\circ h=h $$ showing $f’:=g=h$ as needed.  Conversely, follows from the statements above.

Theorem. Let $f:X\to Y$ be a function.  If $f$ and $g:Y\to Z$ are bijections, then $g\circ f$ is a bijection and  $$ (g\circ f)^{-1}=f^{-1}\circ g^{-1}. $$

Proof. If $f$ and $f$ are bijections, then $g\circ f$ is a bijection and  $(g\circ f)^{-1}=f^{-1}\circ g^{-1}$. Therefore, the inverse of a function $f:X\to Y$ is a function $f^{-1}:Y\to X$ if and only if $f$ is a bijection. Since $f$ and $g$ are bijections, $f^{-1}:Y\to X$ and $g^{-1}:Z\to Y$ are functions. Hence, $f^{-1}\circ g^{-1}:Z\to X$ is a function.  It follows that, $g\circ f$ is injective and surjective, and so a bijection. Thus $(g\circ f)^{-1}:Z\to X$ is also a function. Let $z\in Z$. Since $f$ and $g$ are surjections there exists $x\in X$ and $y\in Y$ such that  \begin{equation} \label{surjcomp} g(y)=z  \qquad \text{and} \qquad  f(x)=y,  \end{equation} respectively.  Written in inverse function notation, $y=g^{-1}(z)$ and $x=f^{-1}(y)$.  By substitution, $x=f^{-1}(g^{-1}(z))=(f^{-1}\circ g^{-1})(z)$.  Notice it also follows from \eqref{surjcomp} that $$(g\circ f)(x)=g(f(x))=g(y)=z.$$  Written in inverse function notation we obtain $(g\circ f)^{-1}(z)=x$.