Short Answer TypeConsider f : N → N, g : N → N and h : N → R defined as f (x) = 2 x, g(y) = 3 y + 4 and h(z) = sin z ∀ x, y and z in N. Show that h o (g o f) = (h o g) o f.
Let f, g and h be function from R to R. Show that (f + g) o h = f o h + g o h
(f . g) o h = (f o h) . (g o h)
Long Answer TypeShort Answer TypeLet f : A → B and g : B → C be two functions such that g o f : A → C is defined.
We are given that g o f : A → C is onto. We now prove that g is onto.
Let z ∈C.
Since gof : A → C is onto, so there exists x ∈ A such that (g o f) (x) = z
⇒ g (f (x)) = z
⇒ g (y) = z where y = f(x)
Since a ∈ A and f is a map from A to B ∴ f (x) ∈ B ⇒ y ∈
∴ for given z ∈ C, we have determined y ∈ B such that g (y) = z ∴ g : B → C is onto.
Consider f : {1, 2, 3, 4} → {1, 2, 3, 4} and g : {1, 2, 3, 4} → {1, 2, 3} defined as f (1) = 1, f(2) = 2 , f(3) = f(4) = 3, g (1) = 1, g (2) = 2 and g (3) = g (4) = 3. So, g of is onto but f is not onto.
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