Let g(x + y) = g(x) + g(y) and g(xy) = g(x)g(y). If g(x) is one-to-one and onto, what is g(0)?

Let f : X → Y and g : Y → X. If g ◦ f is an identity function on X and f ◦ g is an identityfunction on Y , then, show that(a) f is one-one(b) f is onto(c) g = f −1

Let X, Y and Z be any non-empty sets and let f and g be one-one functions of X onto Y andY onto Z respectively so that f and g are both invertible. Then, show that(a) g ◦ f is one-one(b) g ◦ f is onto(c) (g ◦ f )−1 = f −1 ◦g−1

f: R+→R defined by f (x)=2x, x∈ (0,1), f (x) = 3x, x∈ [1,∞) is one -one, onto neither one-one nor onto  one-one, not onto onto

4. Determine whether each of these functions from {a, b, c, d} to itself is one-to-one (onto)a) f (a) = b, f (b) = a, f (c) = c, f (d) = db) f (a) = b, f (b) = b, f (c) = d, f (d) = cc) f (a) = d, f (b) = b, f (c) = c, f (d) = d

What is g(f(x)) if f(x) = g-1 (x)?