Let A = {a, b c}, B = {u, v, w} and let f and g be two functions from

1.	Let A = {a, b c}, B = {u, v, w} and let f and g be two functions from A to B and from B to A respectively defined as: f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)}.Show that f and g both are bijections and find fog and gof.
Answer» Given, A = {a, b, c}, B = {u, v, w} and f = A → B and g: B → A defined by f = {(a, v), (b, u), (c, w)} and g = {(u, b), (v, a), (w, c)} For both f and g, different elements of domain have different images ∴ f and g are one – one Again, for each element in co – domain of f and g, there is a pre – image in the domain ∴ f and g are onto Thus, f and g are bijective. Now, gof = {(a, a), (b, b), (c, c^.)} and fog = {(u, u), (v, v), (w, w)}

Let A = {a, b c}, B = {u, v, w} and let f and g be two functions from A to B and from B to A respectively defined as: f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)}.Show that f and g both are bijections and find fog and gof.

Answer»

Given, A = {a, b, c}, B = {u, v, w} and

f = A → B and g: B → A defined by

f = {(a, v), (b, u), (c, w)} and

g = {(u, b), (v, a), (w, c)}

For both f and g, different elements of domain have different

images

∴ f and g are one – one

Again, for each element in co – domain of f and g, there is a pre – image in the domain

∴ f and g are onto

Thus, f and g are bijective.

Now,

gof = {(a, a), (b, b), (c, c^.)} and

fog = {(u, u), (v, v), (w, w)}