Question

Let A = {1, 3, 4, 7}, B = {u, v w, x, y, z} and f : A → B such that 1 → u, 3 → v, 7 → x. Is f a function ?​

Answer 1

Step-by-step explanation:

(i) Test of bijection of f:

f={a,v),(b,u),(c,w)} and f:A→B

Injectivity of f: We can see, no two elements of A have the same image in B.

∴ f is injective function.

Surjectivity of f: Co-domain of f={u,v,w}

Range of f={u,v,w}

Both are same.

∴ f is surjective function.

Since, f injective and surjective function then, it is a bijection.

(ii) Test of bijection of g:

⇒ g{(u,b),(v,a),(w,c)} and g:B→A

Injectivity of g: No two elements of B have the same image in A.

∴ g is injective function.

Surjectivity of g: Co-domain of g={a,b,c}

Range of g={a,b,c}

Both are the same.

∴ g is surjecutve function.

Since, g injective and surjective function then, it is a bijection.

(iii) Finding f∘g:

Co-domain of g is same as the domain of f.

So, f∘g exists and f∘g:{u,v,w}→{u,v,w}

⇒ (f∘g)(u)=f[g(u)]=f(b)=u

⇒ (f∘g)(v)=f[g(v)]=f(a)=v

⇒ (f∘g)(w)=f[g(w)]=f(c)=w

So, f∘g={(u,u),(v,v),(w,w)}

(iv) Finding g∘f:

Co-domain of f is same as the domain of g.

So, f∘g exists and g∘f:{a,b,c}→{a,b,c}

⇒ (g∘f)(a)=g[f(a)]=g(v)=a

⇒ (g∘f)(b)=g[f(b)]=g(u)=b

⇒ (g∘f)(c)=g[f(c)]=g(w)=c

∴ g∘f={(a,a),(b,b),(c,c)}

Answer 2

Step-by-step explanation:

Since for every element of co-domain set B there is one and only one pre-image in the domain set A, the function is injective as well as surjective. Hence the function is bijective.

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