Consider f:1,2,3→ {a,b,c}given by f(1)=a, f(2)=b and f(3)=c. Find the (f−1)−1 of (f−1) . Show that (f−1)−1=f.
Consider f:1,2,3→ {a,b,c}given by f(1)=a, f(2)=b and f(3)=c. Find the (f−1)−1 of (f−1) . Show that (f−1)−1=f.
Here, function f:[1,2,3]→ {a,b,c}is given by,
f(1)=a, f(2)=b and f(3)=c
if we define g:{a,b,c} → [1,2,3]as g(a)=1, g(b)=2, g(c)=3, then we have
(fog)(a)=f(g(a))=f(1)=a
(fog)(b)=f(g(b))=f(2)=b
(fog)(c)=f(g(c))=f(3)=c
and (gof)(1)=g(f(1))=f(a)=1
(gof)(2)=g(f(2))=f(b)=2
(gof)(3)=g(f(3))=f(c)=3
Therefore, gof=Ix and fog=Iy, where X ={1,2,3} and Y={a,b,c}
Thus, the inverse of f exists and f−1=g.
Therefore, f−1:{a,b,c} → {1,2,3} is given by,
f1(a)=1,f−1(b)=2,f−1(C)=3
Let us now find the inverse of f−1 i.e., find the inverse of g.
If we define h:{1,2,3} → {a,b,c}as
h(1)=a, h(2)=b, h(3)=c, then we have
(goh)(1)=g(h(1))=g(a)=1
(goh)(2)=g(h(2))=g(b)=2
(goh)(3)=g(h(3))=g(c)=3
and (hog)(a)=h(g(a))=h(1)=a
(hog)(b)=h(g(b))=h(2)=b
(hog)(c)=h(g(c))=h(3)=c
Therefore, goh=IX and hog =IY, where X ={1,2,3}and Y ={a,b,c},
Thus, the inverse of g exists and g−1=h⇒(f−1)−1=h.
It can be noted that h =f. Hence, (f−1)−1=f.
Here, function f:[1,2,3]→ {a,b,c}is given by,
f(1)=a, f(2)=b and f(3)=c
if we define g:{a,b,c} → [1,2,3]as g(a)=1, g(b)=2, g(c)=3, then we have
(fog)(a)=f(g(a))=f(1)=a
(fog)(b)=f(g(b))=f(2)=b
(fog)(c)=f(g(c))=f(3)=c
and (gof)(1)=g(f(1))=f(a)=1
(gof)(2)=g(f(2))=f(b)=2
(gof)(3)=g(f(3))=f(c)=3
Therefore, gof=Ix and fog=Iy, where X ={1,2,3} and Y={a,b,c}
Thus, the inverse of f exists and f−1=g.
Therefore, f−1:{a,b,c} → {1,2,3} is given by,
f1(a)=1,f−1(b)=2,f−1(C)=3
Let us now find the inverse of f−1 i.e., find the inverse of g.
If we define h:{1,2,3} → {a,b,c}as
h(1)=a, h(2)=b, h(3)=c, then we have
(goh)(1)=g(h(1))=g(a)=1
(goh)(2)=g(h(2))=g(b)=2
(goh)(3)=g(h(3))=g(c)=3
and (hog)(a)=h(g(a))=h(1)=a
(hog)(b)=h(g(b))=h(2)=b
(hog)(c)=h(g(c))=h(3)=c
Therefore, goh=IX and hog =IY, where X ={1,2,3}and Y ={a,b,c},
Thus, the inverse of g exists and g−1=h⇒(f−1)−1=h.
It can be noted that h =f. Hence, (f−1)−1=f.
