(a) Let us show that the function f:S→S defined by f(a1a2a3a4)=a4a3a2a1 is a bijection.
Let f(a1a2a3a4)=f(b1b2b3b4). Then a4a3a2a1=b4b3b2b1, and hence a4=b4,a3=b3,a2=b2,a1=b1. It follows that a1a2a3a4=b1b2b3b4, and consequently, f is injective. For any b1b2b3b4∈S we have that f(b4b3b2b1)=b1b2b3b4, and hence f is surjective.
(b) Let us show that f−1=f, that is f−1(a1a2a3a4)=a4a3a2a1. Indeed,
f∘f−1(a1a2a3a4)=f(a4a3a2a1)=a1a2a3a4 and
f−1∘f(a1a2a3a4)=f−1(a4a3a2a1)=a1a2a3a4.
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