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# Let $A$ and $B$be sets. Show that$f : A × B \to\; B×A$such that $f(a,b)\,=\,(b,a)$ is bijective function

Toolbox:
• A function $f:A \to B$ where for every $x_1,x_2 \in X,f(x_1)=f(x_2)⇒x_1=x_2$ is called a one-one or injective function.
• A function $f:X \to Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x)=y.$
Given: $f : A × B \to\; B×A$ such that $f(a,b)\,=\,(b,a)$

Step1 - Injective or one-one

Consider two elements $x_1= (a,b)$ and and $x_2= (c,d)$

$f(x_1) =f(x_2) => f(a,b) = f(c,d)$

This implies that $(b,a) =(d,c)$

This is possible only if a=c and b=d

$=> x_1=x_2$

hence the function f defined by $f : A × B \to\; B×A$ such that $f(a,b)\,=\,(b,a)$ is one-one

Step2- Surjective or onto

Let the fuction f be defined by $f : A × B \to\; B×A$such that $f(a,b)\,=\,(b,a)$

the function is onto if for every element $y= (b,a) \in B \times A$ there exist an element $x=(a,b) \in A \times B$ such that $f(x) = y$

This is always possible for all $a\in A , and , b\in B$

Hence the function f is onto

Solution- The function f defined by $f : A × B \to\; B×A$such that $f(a,b)\,=\,(b,a)$ is both one one and onto

edited Mar 19, 2013 by meena.p