In the previous section,
we saw two solved examples related to invertible functions. We saw composite of composite functions also. In
this section, we will see two more solved examples. We will also see a new property of invertible functions.
Solved example 17.27
Consider two functions:
f: {1,2,3}→{a,b,c} defined as:
f(1) = a, f(2) = b, f(3) = c
g: {a,b,c}→{apple, ball, cat} defined as:
g(a) = apple, g(b) = ball, g(c) = cat.
I. Show that, f, g and (g∘f) are invertible.
II. Find out f-1, g-1 and (g∘f)-1.
III. Show that (g∘f)-1 = f-1∘g-1
Solution:
Part (I): Proving that f, g and (g∘f) are invertible.
1. f is a function. Domain is {1,2,3} and codomain is {a,b,c}.
• We know that f is a set. The elements of this set are ordered pairs of the form (x,y).
•
Based on the given data, we can write all the ordered pairs in f:
f = {(1,a), (2,b), (3,c)}
•
We see that:
♦ Each element in the domain has a unique image. So f is one-one.
♦ All elements in the codomain are images. So f is onto.
•
Since f is both one-one and onto, it is invertible.
2. g is a function. Domain is {a,b,c} and codomain is {apple, ball, cat}.
• We know that g is a set. The elements of this set are ordered pairs of the form (x,y).
•
Based on the given data, we can write all the ordered pairs in g:
g = {(a,apple), (b,ball), (c,cat)}
•
We see that:
♦ Each element in the domain has a unique image. So g is one-one.
♦ All elements in the codomain are images. So g is onto.
•
Since g is both one-one and onto, it is invertible.
3. (g∘f)(x) = g(f(x))
•
For (g∘f),
♦ domain is the domain of f which is {1,2,3}
♦ codomain is the codomain of g which is {apple, ball, cat}
•
So we get:
(g∘f)(1) = g(f(1)) = g(a) = apple
(g∘f)(2) = g(f(2)) = g(b) = ball
(g∘f)(3) = g(f(3)) = g(c) = cat
• We know that (g∘f) is a set. The elements of this set are ordered pairs of the form (x,y).
•
Based on the above results, we can write all the ordered pairs in (g∘f):
(g∘f) = {(1,apple), (2,ball), (3,cat)}
•
We see that:
♦ Each element in the domain has a unique image. So (g∘f) is one-one.
♦ All elements in the codomain are images. So (g∘f) is onto.
•
Since (g∘f) is both one-one and onto, it is invertible.
Part (II): Finding f-1, g-1 and (g∘f)-1.
1. Finding f-1:
(i) In part (I), we saw that:
f = {(1,a), (2,b), (3,c)}
• So we can write the reverse of f. We will call it p.
• Since f is one-one and onto, the reverse can be easily written:
p = {(a,1), (b,2), (c,3)}
• The domain of p is {a,b,c}. And codomain is {1,2,3}
(ii) Now we have both f and p. We can calculate (p∘f)
(p∘f) means, output of f is used as the input of p.
• The two functions are:
f: {1,2,3}→{a,b,c}
p: {a,b,c}→{1,2,3}
• We know that, (p∘f) directly connects the domain of f to the codomain of p. So (p∘f) is from {1,2,3} to {1,2,3}.
• When the input is 1, we get:
(p∘f)(1) = p(f(1)) = p(a) = 1
• When the input is 2, we get:
(p∘f)(2) = p(f(2)) = p(b) = 2
• When the input is 3, we get:
(p∘f)(3) = p(f(3)) = p(c) = 3
• So whatever input we give, the output will be the same. That means, (p∘f) is an identity function.
• The inputs are taken from {1,2,3}. So we can write:
(p∘f) = I{1,2,3}.
(iii) Similarly, we can calculate (f∘p)
(f∘p) means, output of p is used as the input of f.
• The two functions are:
f: {1,2,3}→{a,b,c}
p: {a,b,c}→{1,2,3}
• We know that, (f∘p) directly connects the domain of p to the codomain of f. So (f∘p) is from {a,b,c} to {a,b,c}.
• When the input is a, we get:
(f∘p)(a) = f(p(a)) = f(1) = a
• When the input is 2, we get:
(f∘p)(b) = f(p(b)) = f(2) = b
• When the input is 3, we get:
(f∘p)(c) = f(p(c)) = f(3) = c
• So whatever input we give, the output will be the same. That means, (f∘p) is an identity function.
• The inputs are taken from {a,b,c}. So we can write:
(f∘p) = I{a,b,c}.
(iv). Let us write a summary:
• We are given a function f: {1,2,3}→{a,b,c}
• We wrote the reverse function p: {a,b,c}→{1,2,3}
• From (ii), we got: (p∘f) = I{1,2,3}
• From (iii), we got: (f∘p) = I{a,b,c}
• So f is invertible.
• Also, the inverse of f = f-1 = p
• That means:
The inverse of f: {1,2,3}→{a,b,c} is:
p: {a,b,c}→{1,2,3}
2. Finding g-1:
(i) In part (I), we saw that:
g = {(a,apple), (b,ball), (c,cat)}
• So we can write the reverse of g. We will call it q.
• Since g is one-one and onto, the reverse can be easily written:
q = {(apple,a), (ball,b), (cat,c)}
• The domain of q is {apple, ball, cat}. And codomain is {a,b,c}
(ii) Now we have both g and q. We can calculate (q∘g)
(q∘g) means, output of g is used as the input of q.
• The two functions are:
g: {a,b,c}→{apple, ball, cat}
q: {apple, ball, cat}→{a,b,c}
• We know that, (q∘g) directly connects the domain of g to the codomain of q. So (q∘g) is from {a,b,c} to {a,b,c}.
• When the input is a, we get:
(q∘g)(a) = q(g(a)) = q(apple) = a
• When the input is b, we get:
(q∘g)(b) = q(g(b)) = q(ball) = b
• When the input is c, we get:
(q∘g)(c) = q(g(c)) = q(cat) = c
• So whatever input we give, the output will be the same. That means, (q∘g) is an identity function.
• The inputs are taken from {a,b,c}. So we can write:
(q∘g) = I{a,b,c}.
(iii) Similarly, we can calculate (g∘q)
(g∘q) means, output of q is used as the input of g.
• The two functions are:
g: {a,b,c}→{apple, ball, cat}
q: {apple, ball, cat}→{a,b,c}
• We know that, (g∘q) directly connects the domain of q to the codomain of g. So (g∘q) is from {apple, ball, cat} to {apple, ball, cat}.
• When the input is apple, we get:
(g∘q)(apple) = g(q(apple)) = g(a) = apple
• When the input is 2, we get:
(g∘q)(ball) = g(q(ball)) = g(b) = ball
• When the input is 3, we get:
(g∘q)(cat) = g(q(cat)) = g(c) = cat
• So whatever input we give, the output will be the same. That means, (g∘q) is an identity function.
• The inputs are taken from {apple, ball, cat}. So we can write:
(g∘q) = I{apple, ball, cat}.
(iv). Let us write a summary:
• We are given a function g: {a,b,c}→{apple, ball, cat}
• We wrote the reverse function q: {apple, ball, cat}→{a,b,c}
• From (ii), we got: (q∘g) = I{a,b,c}
• From (iii), we got: (g∘q) = I{apple, ball, cat}
• So g is invertible.
• Also, the inverse of g = g-1 = q
• That means:
The inverse of g: {a,b,c}→{apple, ball, cat} is:
q: {apple, ball, cat}→{a,b,c}
3. Finding (g∘f)-1:
(i) In part (I), we saw that:
(g∘f) = {(1,apple), (2,ball), (3,cat)}
• So we can write the reverse of (g∘f). We will call it r.
• Since (g∘f) is one-one and onto, the reverse can be easily written:
r = {(apple,1), (ball,2), (cat,3)}
• The domain of r is {apple, ball, cat}. And codomain is {1,2,3}
(ii) Now we have both (g∘f) and r. We can calculate (r∘(g∘f))
(r∘(g∘f)) means, output of (g∘f) is used as the input of r.
• The two functions are:
(g∘f): {1,2,3}→{apple, ball, cat}
r: {apple, ball, cat}→{1,2,3}
• We know that, (r∘(g∘f)) directly connects the domain of (g∘f) to the codomain of r. So (r∘(g∘f)) is from {1,2,3} to {1,2,3}.
• When the input is 1, we get:
(r∘(g∘f))(1) = r((g∘f)(1)) = r(apple) = 1
• When the input is 2, we get:
(r∘(g∘f))(2) = r((g∘f)(2)) = r(ball) = 2
• When the input is 3, we get:
(r∘(g∘f))(3) = r((g∘f)(3)) = r(cat) = 3
• So whatever input we give, the output will be the same. That means, (r∘(g∘f)) is an identity function.
• The inputs are taken from {1,2,3}. So we can write:
(r∘(g∘f)) = I{1,2,3}.
(iii) Similarly, we can calculate ((g∘f)∘r)
((g∘f)∘r) means, output of r is used as the input of (g∘f).
• The two functions are:
(g∘f): {1,2,3}→{apple, ball, cat}
r: {apple, ball, cat}→{1,2,3}
• We know that, ((g∘f)∘r) directly connects the domain of r to the codomain of (g∘f). So ((g∘f)∘r) is from {apple, ball, cat} to {apple, ball, cat}.
• When the input is apple, we get:
((g∘f)∘r)(apple) = (g∘f)(r(apple)) = (g∘f)(1) = apple
• When the input is 2, we get:
((g∘f)∘r)(ball) = (g∘f)(r(ball)) = (g∘f)(2) = ball
• When the input is 3, we get:
((g∘f)∘r)(cat) = (g∘f)(r(cat)) = (g∘f)(3) = cat
• So whatever input we give, the output will be the same. That means, ((g∘f)∘r) is an identity function.
• The inputs are taken from {apple, ball, cat}. So we can write:
((g∘f)∘r) = I{apple, ball, cat}.
(iv). Let us write a summary:
• We are given a function (g∘f): {1,2,3}→{apple, ball, cat}
• We wrote the reverse function r: {apple, ball, cat}→{1,2,3}
• From (ii), we got: (r∘(g∘f)) = I{1,2,3}
• From (iii), we got: ((g∘f)∘r) = I{apple, ball, cat}
• So (g∘f) is invertible.
• Also, the inverse of (g∘f) = (g∘f)-1 = r
• That means:
The inverse of (g∘f): {1,2,3}→{apple, ball, cat} is:
r: {apple, ball, cat}→{1,2,3}
Part III: Showing that, (g∘f)-1 = f-1∘g-1.
1. From part (II)(1), we have:
f-1 = p
Where p: {a,b,c}→{1,2,3}
• In set form, we can write:
f-1 = {(a,1), (b,2), (c,3)}
2. From part (II)(2), we have:
g-1 = q
Where q: {apple, ball, cat}→{a,b,c}
• In set form, we can write:
q-1 = {(apple,a), (ball,b), (cat,c)}
3. From part (II)(3), we have:
(g∘f)-1 = r
Where r: {apple, ball, cat}→{1,2,3}
• In set form, we can write:
(g∘f)-1 = {(apple,1), (ball,2), (cat,3)}
4. Now we can calculate f-1∘g-1.
(f-1∘g-1)(x) = f-1(g-1(x))
•
For (f-1∘g-1),
♦ domain is the domain of g-1 which is {apple, ball, cat}
♦ codomain is the codomain of f-1 which is {1,2,3}
•
So we get:
(f-1∘g-1)(apple) = f-1(g-1(apple)) = f-1(a) = 1
(f-1∘g-1)(ball) = f-1(g-1(ball)) = f-1(b) = 2
(f-1∘g-1)(cat) = f-1(g-1(cat)) = f-1(c) = 3
5. We can write the result in (4), in the form of a set:
(f-1∘g-1) = {(apple,1), (ball,2), (cat,3)}
6. Comparing the results in (3) and (5), we get:
(g∘f)-1 = (f-1∘g-1)
The result obtained in the above solved example 17.27 can be used in general. We can write:
If f: X→Y and g: Y→Z are two invertible functions, then (g∘f) is also invertible. The inverse of (g∘f) is (f-1∘g-1)
The proof can be written in 7 steps:
1. First we will write some basic details:
(i) Given that, f: X→Y is an invertible function.
• So we can write:
♦ f(x) = y
♦ f-1(y) = x
(ii) Given that, g: Y→Z is an invertible function.
• So we can write:
♦ g(y) = z
♦ g-1(z) = y
(iii) Note that, domain of g is same as codomain of f.
2. Now we recall an important property of inverse functions.
• If the inverse of f is f-1, then two conditions will be satisfied:
(i) (f-1∘f) = IX
(ii) (f∘f-1) = IY
Where
♦ X is the domain of f
♦ Y is the domain of f-1
• Similarly, if the inverse of g is g-1, then two conditions will be satisfied:
(i) (g-1∘g) = IY
(ii) (g∘g-1) = IZ
Where
♦ Y is the domain of g
♦ Z is the domain of g-1
3. Next step is to determine the domain and codomain of the composite functions.
(i) Domain of (g∘f):
• (g∘f) connects the domain of f with the codomain of g.
♦ So the domain of (g∘f) is the domain of f, which is X.
♦ Also the codomain of (g∘f) is the codomain of g, which is Y.
(ii) Domain of (f-1∘g-1):
• (f-1∘g-1) connects the domain of g-1 with the codomain of f-1.
♦ So the domain of (f-1∘g-1) is the domain of g-1, which is Z.
♦ Also the codomain of (f-1∘g-1) is the codomain of f-1, which is X.
4. Now we can use the property mentioned in step (2).
• That is:
If the inverse of (g∘f) is (f-1∘g-1), then two conditions will be satisfied:
(i) (f-1∘g-1)∘(g∘f) = IX
(ii) (g∘f)∘(f-1∘g-1) = IZ
Where
♦ X is the domain of (g∘f)
♦ Z is the domain of (f-1∘g-1)
5. First we show that (f-1∘g-1)∘(g∘f) = IX.
$\begin{array}{ll}{} &{(f^{-1} \circ g^{-1})(g \circ f)} & {~=~} &{\Bigl((f^{-1} \circ g^{-1}) \circ g \Bigr) \circ f ~\color {magenta}{\text{- - - (A)}}} &{} \\
{} &{} & {~=~} &{\Bigl(f^{-1} \circ (g^{-1} \circ g) \Bigr) \circ f ~\color {magenta}{\text{- - - (B)}}} &{} \\
{} &{} & {~=~} &{(f^{-1} \circ I_{Y}) \circ f ~\color {magenta}{\text{- - - (C)}}} &{} \\
{} &{} & {~=~} &{I_X ~\color {magenta}{\text{- - - (D)}}} &{} \\
\end{array}
$
◼ Remarks:
• Line marked as A:
Here we use the formula: h∘(g∘f) = (h∘g)∘f
• Line marked as B:
Here also, we use the formula: h∘(g∘f) = (h∘g)∘f
• Line marked as C:
Here we use the fact that: (g-1∘g) = IY
• Line marked as D:
We must simplify (f-1∘IY)∘f. This can be done in two steps:
(i) First we will determine (f-1∘IY)
• Here, the output of IY is used as the input for f-1.
• Also this function connects the domain of IY with the codomain of f-1.
♦ Domain of IY is Y
♦ Codomain of f-1 is X
So we get: (f-1∘IY) = f-1(IY) = f-1(y) = x.
(ii) Now we can determine (f-1∘IY)∘f
• Here, the output of f is used as the input for (f-1∘IY).
• Also this function connects the domain of f with the codomain of (f-1∘IY).
♦ Domain of f is X
♦ Codomain of (f-1∘IY) is also X
So we get: (f-1∘IY)∘f = IX.
6. In a similar way, we can show that:
(g∘f)∘(f-1∘g-1) = IZ.
7. So both conditions mentioned in (4) are satisfied.
• Therefore it is proved that:
If f: X→Y and g: Y→Z are two invertible functions, then (g∘f) is also invertible. The inverse of (g∘f) is (f-1∘g-1)
Solved example 17.28
Let S = {1,2,3}. Determine whether the functions f: S→S defined as below have inverses. Find f-1 if it exists.
(a) f = {(1,1), (2,2), (3,3)}
(b) f = {(1,2), (2,1), (3,1)}
(c) f = {(1,3), (3,2), (2,1)}
Solution:
Part (a): f = {(1,1), (2,2), (3,3)}
• Here f is both one-one and onto. So f-1 exists.
• We can write: f-1 = {(1,1), (2,2), (3,3)}
Part (b): f = {(1,2), (2,1), (3,1)}
• Here f(2) = f(3) = 1.
So f is not one-one and onto. Therefore, f-1 does not exist.
Part (c): f = {(1,3), (3,2), (2,1)}
• Here f is both one-one and onto. So f-1 exists.
• We can write: f-1 = {(3,1), (2,3), (1,2)}
Link to a few more solved examples is given below:
In the next section, we will see binary operations.
No comments:
Post a Comment