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

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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})\\

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