**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.