Hi everyone,

I've just encountered a tricky inverse question. Can someone please give me a hint.

Assume that f and g are both one-to-one functions. Express (f of g )^(-1) in terms of f^(-1) and g^(-1).

Thanks a lot everyone.

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- Nov 20th 2011, 09:28 PMpohkidTricky inverse question
Hi everyone,

I've just encountered a tricky inverse question. Can someone please give me a hint.

Assume that f and g are both one-to-one functions. Express (f of g )^(-1) in terms of f^(-1) and g^(-1).

Thanks a lot everyone. - Nov 21st 2011, 03:36 AMAmerRe: Tricky inverse question
you should know that for any function f ,f^-1 of f has the property f(f^-1)(x) = x or f^-1(f(x)) = x

so suppose that "h " is the inverse function of $\displaystyle (f\;o\;g)$ now just find "h" such that

$\displaystyle (f\; o\; g\;o\; h )(x) = x \Rightarrow \;\;f(g(h(x))) =x $

can you handle it ?

take inverse function of "f" then for "g" - Nov 21st 2011, 04:39 AMPlatoRe: Tricky inverse question
Suppose that $\displaystyle (x,z) \in f \circ g$ then $\displaystyle \left( {\exists y} \right)\left[ {(x,y) \in g\;\& \,(y,z) \in f} \right]$.

So $\displaystyle (y,x) \in g^{ - 1} \;\& \,(z,y) \in f^{ - 1} $ or $\displaystyle \left( {z,x} \right) \in g^{ - 1} \circ f^{ - 1} $.

Thus $\displaystyle \left( {f \circ g} \right)^{ = 1} =~?$ - Nov 21st 2011, 06:24 AMDevenoRe: Tricky inverse question
another way to look at this is:

g:x-->y f:y-->z

fog:x-->y-->z, or just fog:x-->z for short.

1-1 means we can "reverse the arrows" (because each y = g(x) came from a single x, and each z = f(y) came from a single y). so:

f^-1:z-->y g^-1:y-->x

now express a function that takes z-->x in 2 different ways: one as the inverse of fog, and another as some composition......