Mathematical Immaturity

2.8 Exercises

2.$\quad$ Let $V = \{0, 1, 2\}.$ Describe all functions $T: V \rightarrow V$ for which $T(V) = V.$ There are six altogether. Label them as $T_1, \ldots, T_6$ 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 six functions $T: V \to V$ for which $T(V) = V$ are: \begin{align*} T_1 &= \{(0, 0), (1, 1), (2, 2)\} \\ T_2 &= \{(0, 0), (1, 2), (2, 1)\} \\ T_3 &= \{(0, 1), (1, 0), (2, 2)\} \\ T_4 &= \{(0, 1), (1, 2), (2, 0)\} \\ T_5 &= \{(0, 2), (1, 1), (2, 0)\} \\ T_6 &= \{(0, 2), (1, 0), (2, 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 | c | c } & T_1 & T_2 & T_3 & T_4 & T_5 & T_6 \\ \hline T_1 & T_1T_1 = T_1 & T_1T_2 = T_2 & T_1T_3 = T_3 & T_1T_4 = T_4 & T_1T_5 = T_5 & T_1T_6 = T_6 \\ \hline T_2 & T_2T_1 = T_2 & T_2T_2 = T_1 & T_2T_3 = T_6 & T_2T_4 = T_5 & T_2T_5 = T_4 & T_2T_6 = T_3 \\ \hline T_3 & T_3T_1 = T_3 & T_3T_2 = T_4 & T_3T_3 = T_1 & T_3T_4 = T_2 & T_3T_5 = T_6 & T_3T_6 = T_5 \\ \hline T_4 & T_4T_1 = T_4 & T_4T_2 = T_3 & T_4T_3 = T_5 & T_4T_4 = T_6 & T_4T_5 = T_2 & T_4T_6 = T_1 \\ \hline T_5 & T_5T_1 = T_5 & T_5T_2 = T_6 & T_5T_3 = T_4 & T_5T_4 = T_3 & T_5T_5 = T_1 & T_5T_6 = T_2 \\ \hline T_6 & T_6T_1 = T_6 & T_6T_2 = T_5 & T_6T_3 = T_2 & T_6T_4 = T_1 & T_6T_5 = T_3 & T_6T_6 = T_4 \\ \end{array} All six functions are one-to-one because for each function, $x \neq y$ implies $T(x) \neq T(y).$ Given that $T_1$ is the identity transformation $I_V,$ the inverse of each function $T_i$ is the function $T_j = T^{-1}_i$ that, when composed with $T_i,$ yields $T_1.$ Referring to the above table, we find that \begin{align*} T^{-1}_1 = T_1, \quad T^{-1}_2 = T_2, \quad T^{-1}_3 = T_3, \quad T^{-1}_4 = T_6, \quad T^{-1}_5 = T_5, \quad T^{-1}_6 = T_4 \quad \blacksquare \end{align*}