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

The Answer to the Question
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

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

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

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

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

and hence the function $f$ is one-to-one.