Question -
Answer -
Given f ={(a, v), (b, u), (c, w)}, g = {(u, b),(v, a), (w, c)}.
Also given that A = {a, b, c}, B = {u, v, w}
Now we have to show fand g both are bijective.
Consider f ={(a, v), (b, u), (c, w)} and f: A → B
Injectivity of f: No two elements of A have the same image in B.
So, f is one-one.
Surjectivity of f: Co-domain of f = {u, v, w}
Range of f = {u, v, w}
Both are same.
So, f is onto.
Hence, f is a bijection.
Now consider g ={(u, b), (v, a), (w, c)} and g: B → A
Injectivity of g: No two elements of B have the same imagein A.
So, g is one-one.
Surjectivity of g: Co-domain of g = {a, b, c}
Range of g = {a, b, c}
Both are the same.
So, g is onto.
Hence, g is a bijection.
Now we have to findfog,
we know that Co-domain of g is same as the domain of f.
So, fog exists and fog:{u v, w} → {u, v, w}
(fog) (u) = f (g (u)) = f (b) = u
(fog) (v) = f (g (v)) = f (a) = v
(fog) (w) = f (g (w)) = f (c) = w
So, fog = {(u, u), (v, v), (w, w)}
Now we have to findgof,
Co-domain of f is same as the domain of g.
So, fog exists and gof:{a, b, c} → {a, b, c}
(gof) (a) = g (f (a)) = g (v) = a
(gof) (b) = g (f (b)) = g (u) = b
(gof) (c) = g (f (c)) = g (w) = c
So, gof = {(a, a), (b, b), (c, c)}