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Proof with functions

Q: Let f and g be invertible functions such that:

f: S –> T
g: T –>U
(1) Show g compose f is invertible and,

(2) that (g compose f) inverse = (f inverse) compose (g inverse)

A:  To let you know: I am a very thorough proof write.  This may be lengthy, be forewarned.  However, I am a very good proof writer, so enjoy 🙂

(1)  Goal: to show g compose f is invertible.

For all s ε S, we know that f(s) = t for some t ε T.  And, g(t) = u for some u ε U.  Therefore, for all s ε S there exists a u ε U such that g(f(s)) = u.  Define h: U –> S such h(u) = s.  Therefore, g(f(S)) is invertible and its inverse is h.

(2)  Goal:  to show that (f compose g) inverse is (f inverse) compose (g inverse)

OR to show that h is (f inverse) compose (g inverse) [since h is defined to be (f compose g) inverse]

We define f-1, f inverse, such that if f(s) = t, then s = f-1(t), for all s ε S.

Similarly, g-1, g inverse, such that if g(t) = u, then t = g-1(u), for all t ε T.

Consider g(f(s)) = u for arbitrary s ε S.

Then, g-1(g(f(s))) = g-1(u) –> f(s) = g-1(u).

And, f-1(f(s)) = f-1(g-1(u)) –> s = f-1(g-1(u))

As defined before, we also know h(u) = s, so h(u) = f-1(g-1(u)).

Therefore, h = f inverse composed with g inverse.

Since h is (g compose f) inverse, we have:

(g compose f) inverse is f inverse composed with g inverse.

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