
Remark:
gof is defined only if Rf = Dg.

Example 1:

Suggested answer:


Example 2:
If f : R→R and g: R→R denoted by f(x) = x + 2, and g(x) = 2x2 + 7. Compute fog and gof.
Suggested answer:


Theorem
Statement:
f : S
T, g : T
U and h : U
V be function then
b) If each f and g is one-to-one then so is gof.
c) If each f and g is onto then so is gof.
Proof:
a) (hog)of = ho(gof)
Since, (hog)of and ho(gof) are functions from S to V. We have to prove that for any
the image of S under ho(gof) = image of x under (hog)of.





c) If each f and g is also then so is gof.
To prove gof is onto we have to prove that every element of u Î U is an image element for some x Î S under gof, since g is onto
t Î T such that g(t) = u.
gof(x) = g(f(x)) = g(t) = u

gof : S g U is one-one and onto by (b) and (c).
(gof)-1 exists and (gof)-1 : U®S is also a one-one and onto function.
Again f-1and y-1 exists and one-one and onto because f and g are one-one and onto by (b) and (c).

(
if is one-one and onto)

= f-1 (t) = x ...(ii)
From (i) and (ii), we get
Corollary:
The function f : A g B is one-one and onto.

Proof:
f : A g B and is one-one and onto.
\ for x Î A , we have an element b Î B such that






This proves that f is one-one and onto.
