inverse function from infinite set A to infinite set A...?

Question

GIVEN: A is a infinite Set.
f and g are functions from A to A and f(g(x)) has left and right inverse.
QUESTION: Is it true that from the GIVEN, cannot be concluded that f and g have inverse functions of their own. If so then why ?

Answer 1

first of all, if f°g has a left and a right inverse, then these inverses are equal:

(f°g)° h = id and k°(f°g) = id =>
k = k°(f°g) ° h = h

f°g invertible implies that f is 1-1, and also that g is onto.

now,
(f°g)° h = id
but (f°g)° h = f°(g°h), so f has a right inverse

k°(f°g) = id implies that g has a left inverse:
k°(f°g) = (k°f)°g

Answer 2

f'(x) = f^-a million(x) f(x) = y f^-a million(y) = x d/dx (f^-a million(y)) = d/dx (x) = a million df^-a million/dx * dy/dx = a million df^-a million/dx = f''(x), so f''(x) f'(x) = a million g = f'(x) dg/dx * g = a million g dg = dx a million/2 g^2 = x + C g = sqrt(2x + C) f'(x) = sqrt(2x + C) f(x) = a million/3 (2x+C)^3/2 and that's the respond. Checking: y = a million/3 (2x+C)^3/2 (3y)^2/3 = 2x + C x = a million/2 (3y)^2/3 - C/2 Hmmm - this does not artwork, yet i can not see why. engaged on it... whether it gave me a clue approximately the thank you to try this. The order of f^-a million(n) would be a million/ the order of f(n). So if f(x) is algebraic, the flexibility n would desire to be such that: n - a million/n = a million n^2 - n - a million = 0 n = [a million +/- sqrt(5)]/2 f(x) = x^[(a million+sqrt(5))/2] df/dx = [(a million+sqrt(5))/2] x^[(-a million+sqrt(5))/2] y^[2/(a million+sqrt(5))] = y^[(-a million+sqrt(5))/2] = x