Question #279845

Determine whether each of these functions from {a,b,c,d} to itself is one-to-one. a) f(a)=b, f(b)=a, f(c)=c, f(d)=d


Expert's answer

Let us determine whether the function from {a,b,c,d}\{a,b,c,d\} to itself is one-to-one.

a) f(a)=b,f(b)=a,f(c)=c,f(d)=d.f(a)=b, f(b)=a, f(c)=c, f(d)=d.


Taking into account that preimage of each point is singleton: f−1(a)={b}, f−1(b)={a}, f−1(c)={c}, f−1(d)={d},f^{-1}(a)=\{b\},\ f^{-1}(b)=\{a\},\ f^{-1}(c)=\{c\},\ f^{-1}(d)=\{d\},

we conclude that x≠yx\ne y implies f(x)≠f(y)f(x)\ne f(y) for any x,y∈{a,b,c,d},x,y\in\{a,b,c,d\},

and hence the function ff is one-to-one.


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