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 :

Solution:

Proving f is a bijection:

$f=\{(a, v),(b, u),(c, w)\}$ and $f: A \rightarrow B$

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.

Proving g is a bijection:

$g=\{(u, b),(v, a),(w, c)\}$ and $g: B \rightarrow A$

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

Finding  fog:
Co-domain of g is same as the domain of f.

So, $f o g$ exists and $f o g:\{u v, w\} \rightarrow\{u v, w\}$

$(f o g)(u)=f(g(u))=f(b)=u$

$(f o g)(v)=f(g(v))=f(a)=v$

$(f o g)(w)=f(g(w))=f(c)=w$

So, fog $=\{(u, u),(v, v),(w, w)\}$

Finding gof:
Co-domain of f is same as the domain of g.

So, fog exists and $g o f:\{a, b, c\} \rightarrow\{a, b, c\}$

$(g o f)(a)=g(f(a))=g(v)=a$

$(g o f)(b)=g(f(b))=g(u)=b$

$(g o f)(c)=g(f(c))=g(w)=c$

So, gof $=\{(a, a),(b, b),(c, c)\}$