Question

Let$f:G\to G$ be given by, $f\left(a\right)=a^{-1}$. Prove that $f$is a bijection.

Short answer

It is proved that$f$ is a bijection.

Step-by-step solution

Bijective mapping

A mapping $f:A\to B$is said to be bijective if it is both injective and surjective. If

• $x_1\ne x_2$in$A$ implies $f\left(x_1\right)\ne f\left(x_2\right)$in$B$ .
• $f\left(A\right)=B$.

To prove injective

Let $a,b\in G$such that,$f\left(a\right)=f\left(b\right)$

.$\Rightarrow a^{-1}=b^{-1}$

Multiplying by from the right and then by $b$from the left, we get,

$\begin{array}{l}a\left(a^{-1}\right)b=a\left(b^{-1}\right)b\\ \Rightarrow \left(a{a}^{-1}\right)b=a\left(b^{-1}b\right)\\ \Rightarrow eb=ae\\ \Rightarrow a=b\end{array}$

which shows that,$f$ is injective.

Final conclusion

Let$b\in G$ be arbitrary.Now, we will show that $\exists a\in G$, such that,

$f\left(a\right)=b$

.$\Rightarrow a^{-1}=b$ Now, multiplying by $b^{-1}$from the left and by from the right we get,

$\begin{array}{l}a\left(a^{-1}\right)b^{-1}=a\left(b\right)b^{-1}\\ \Rightarrow \left(a{a}^{-1}\right)b^{-1}=a\left(b{b}^{-1}\right)\\ \Rightarrow b^{-1}=a\end{array}$

It shows that$f$ is surjective.

Hence,$f$ is a bijection.