Question

If f:x->Y is bijective and g:y->z is bijective, then prove that (gof )^(-1)=f^(-1)og^(-1).

Answer 1

To prove that (gof)^(-1) = f^(-1)og^(-1), we need to show that applying the composition of f and g followed by its inverse gives the same result as applying the inverses of f and g in the reverse order.

Step-by-step explanation:

To prove that (gof)^(-1) = f^(-1)og^(-1), we need to show that applying the composition of f and g followed by its inverse gives the same result as applying the inverses of f and g in the reverse order.

First, let's consider (gof)^(-1). Applying (gof)^(-1) to a value z, we get:

(gof)^(-1)(z) = f^(-1)(g^(-1)(z))

On the other hand, applying f^(-1)og^(-1) to z, we get:

f^(-1)og^(-1)(z) = f^(-1)(g^(-1)(z))

Since (gof)^(-1)(z) = f^(-1)og^(-1)(z) for all values of z, we can conclude that (gof)^(-1) = f^(-1)og^(-1).