(i) f1 = {(1, 3), (2, 5), (3, 7)} ; A = {1, 2, 3}, B = {3, 5, 7}
Injectivity:
f1 (1) = 3
f1 (2) = 5
f1 (3) = 7
Every element of A has different images in B.
So, f1 is one-one.
Surjectivity:
Co-domain of f1 = {3, 5, 7}
Range of f1 =set of images = {3, 5, 7}
Co-domain = range
So, f1 is onto.
(ii) f2 = {(2, a), (3, b), (4, c)} ; A = {2, 3, 4}, B = {a, b, c}
Injectivity:
f2 (2) = a
f2 (3) = b
f2 (4) = c
Every element of A has different images in B.
So, f2 is one-one.
Surjectivity:
Co-domain of f2 = {a, b, c}
Range of f2 = set of images = {a, b, c}
Co-domain = range
So, f2 is onto.
(iii) f3 = {(a, x), (b, x), (c, z), (d, z)} ; A = {a, b, c, d,}, B = {x, y, z}
Injectivity:
f3 (a) = x
f3 (b) = x
f3 (c) = z
f3 (d) = z
a and b have the same image x. (Also c and d have the same image z)
So, f3 is not one-one.
Surjectivity:
Co-domain of f1 ={x, y, z}
Range of f1 =set of images = {x, z}
So, the co-domain is not same as the range.
So, f3 is not onto.