Question

if f:AtoB,g:BtoC are bijection then prove that gof:a to c is a bijection​

Answer 1

To show a function is bijective:

• Show it's one-one
• Show it's onto.

Step-by-step explanation:

Showing it's one - one function:

Consider

f(x) is one-one and onto function,

g(x) in one-one and onto function,

Let ( $\\\tt x_1$, $\\\tt x_2$ )∈ A now,

f($\\\tt x_1$) = f($\\\tt x_2$)

$\\\tt x_1 = x_2$  (since f(x)is one - one)  -------(i)

similarly for g(x)

Let ( $\\\tt x_1$, $\\\tt x_2$ )∈ A now,

g($\\\tt x_1$) = g($\\\tt x_2$)

$\\\tt x_1 = x_2$  (since g(x)is one - one too) ----------(ii)

Now for

gof($\\\tt x_1$) = gof($\\\tt x_2$)

g(f($\\\tt x_1$)) = g(f($\\\tt x_2$))

f($\\\tt x_1$) = f($\\\tt x_2$) (from ii)

$\\\tt x_1$= $\\\tt x_2$ (from i)

therefore gof is one- one function.

Now to show gof is onto.

g(f(x))

Let f(x) = E

g(f(x)) = g(E) = C

since g(x) is onto function.

Therefore fog is bijective.