- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.8 Exercises
1. $\quad$ Let $V = \{0, 1\}.$ Describe all functions $T: V \rightarrow V.$ There are four altogether. Label them as $T_1, T_2, T_3, T_4$ and make a multiplication table showing the composition of each pair. Indicate which functions are one-to-one on $V$ and give their inverses.
Solution. $\quad$ The four functions that map $V$ onto itself are: \begin{align*} T_1 &= \{(0, 0), (1, 0)\} \\ T_2 &= \{(0, 1), (1, 0)\} \\ T_3 &= \{(0, 0), (1, 1)\} \\ T_4 &= \{(0, 1), (1, 1)\} \end{align*} The composition of the sixteen possible pairs is given in the following multiplication table, where each entry is the composition of the row's function with the column's function. \begin{array}{ c | c | c | c | c } & T_1 & T_2 & T_3 & T_4 \\ \hline T_1 & T_1T_1 = T_1 & T_1T_2 = T_1 & T_1T_3 = T_1 & T_1T_4 = T_1 \\ \hline T_2 & T_2T_1 = T_4 & T_2T_2 = T_3 & T_2T_3 = T_2 & T_2T_4 = T_1 \\ \hline T_3 & T_3T_1 = T_1 & T_3T_2 = T_2 & T_3T_3 = T_3 & T_3T_4 = T_4 \\ \hline T_4 & T_4T_1 = T_4 & T_4T_2 = T_4 & T_4T_3 = T_4 & T_4T_4 = T_4 \\ \end{array} The functions $T_2$ and $T_3$ are one-to-one on $V,$ because for each function, $x \neq y$ implies $T(x) \neq T(y)$ and $T(x) = T(y)$ implies $x = y.$ Then, noting that $T_3$ is the identity function of $V,$ we find that $T_2$ and $T_3$ are their own inverses with $T_2T_2 = T_3 = I_V$ and $T_3T_3 = T_3 = I_V. \quad \blacksquare$