Answer to Question #203451 in Discrete Mathematics for Umair

Question

Suppose f:X→Y and g:Y→Z and both of these are one-to-one and onto. Prove that (g∘f)^(-1) exists and that (g∘f)^(-1)=f^(-1)∘g^(-1).

Answer 1

Let

$f : X \to Y \space and \space g:Y \to Z$

both are one-one and onto functions.

It's mean $gof:X\to Z$ defined.

(1)Prove that

$(gof)^{-1}$ exist:-

we know that if both functions are ono-one , onto and gof defined,

then $(gof)^{-1}$ defined and exist too.

such that

$(gof)^{-1}:Z\to X$

(2)prove that

$(g∘f)^{-1}=f^{-1}∘g^{-1}$ :-

now

$f:X\to Y$ is one-one onto $\implies f^{-1}:Y\to X \space exist.$

$g:Y\to Z$ is one-one onto $\implies g^{-1}:Z\to Y \space exist.$

$thus \space (f^{-1}og^{-1}):Z\to X \space exist$

so domain and codomain of $(gof)^{-1}$ and $(f^{-1}og^{-1})$ are same.

let

$x \isin X, y \isin Y, z\isin Z$

such that

$f(x)=y$ and $g(y)=z$

$(gof)(x)=g[f(x)]=g(y)=z \\$

$\implies (gof)^{-1}(z)=x$ ............(1)

again

$f(x)=y \implies f^{-1}(y)=x......(2)\\ g(y)=z \implies g^{-1}(z)=y......(3)\\$

$\therefore (f^{-1}og^{-1})(z)=f^{-1}[g^{-1}(z)]=f^{-1}(y) \\$

........from equation (3)

$(f^{-1}og^{-1})(z)=x......from\space eq(2)$

thus

from eq (1) and the last eq,

$(gof)^{-1}(z)=(f^{-1}og^{-1})(z)\\ hence\\ (gof)^{-1}=(f^{-1}og^{-1})\\$