## Chapter - 5, Functions

### Section - 5.3 - Inverses of Functions

### Summary

- Suppose . If is one-to-one and onto, then .
- Suppose is a function from to , and suppose that is a function from to . Then and .
- Suppose . Then,
- If there is a function such that then is one-to-one.
- If there is a function such that then is onto.

- Suppose . Then the following statements are equivalent.
- is one-to-one and onto.
- .
- There is a function such that and .

- Suppose , , , and . Then .

**Soln1**

Required function .

**Soln2**

Required function .

**Soln3**

Suppose . Thus we have:

.

.

.

.

Thus we have .

Also, we have .

.

.

.

Thus we have .

Since and , it follows that is one-to-one and onto and .

**Soln4**

Suppose . Clearly is defined for all values of (cube root of negative number is defined). Thus .

Now, .

.

.

.

Thus we have .

Also, .

.

.

.

Thus we have .

Since and , it follows that is one-to-one and onto and .

**Soln5**

Suppose . Clearly .

We have . Thus .

Also, we have . Thus .

Since and , it follows that is one-to-one and onto and .

**Soln6**

Suppose is a function from to . Clearly for any value of . Thus we can say that , where and .

Now, .

.

.

.

.

Thus we have .

Similarly it can be shown that, .

Since and , it follows that is one-to-one and onto and .

Thus **(a)** and **(b)** both are solved.

**Soln7**

**(a)**

To prove , we need to prove that . Suppose . Thus . Since is arbitrary, .

**(b)**

Clearly and . Thus .

**Soln8**

**(a)**

Suppose is an arbitrary element of . Let . Then , so and therefore .

Thus, .

Since was arbitrary, we have shown that , so .

**(b)**

Suppose . Since , it follows . Since , . Thus (using the first half of the proof as required in the solution). It follows .

**Soln9**

Suppose . Suppose such that . Suppose . Thus . It follows that . Thus There exist an element such that and . Thus there exist an element such that . Since is arbitrary, it follows that . Thus is onto.

**Soln10**

()Suppose is an arbitrary pair in . Since , it follows . Since , it follows . Thus for some element , and . Since is a function and , it follows . Thus . It follows . Thus .

() Suppose is an arbitrary pair in . Thus and . Since , it follows . Thus for some element , and . Since is a function and , it follows . Thus . Thus .

Since and , it follows that .

**Soln11**

**(a)**

Since and and , it follows that is onto by the theorem(3.1 in above summary). It is given that is one-to-one. Thus is one-to-one and onto. Thus .

Thus, .

**(b)**

Since and and , it follows that is one-to-one by the theorem(3.2 in above summary). It is given that is onto. Thus is one-to-one and onto. Thus .

Thus, .

**(c)**

Since and and , it follows from theorem(3rd theorem form above summary) that is onto and is one-to-one.

Proof for is not onto:

Suppose is onto. Thus is one-to-one and onto. Thus by theorem, . Applying on both sides, . Since is one-to-one and onto, . Thus . But it contradicts with the given that . Thus we can conclude that is not onto.

Proof for is not one-to-one:

Suppose is one-to-one. Thus is one-to-one and onto. Thus by theorem, . Thus . Since is one-to-one and onto, . Thus . But it contradicts with the given that . Thus we can conclude that is not one-to-one.

**Soln12**

Suppose is one-to-one. Suppose . Clearly . Now we shall prove . Thus we shall do it by proving existence and uniqueness:

*Existence:*

Suppose . Thus . It follows that there exists such that . Thus . Thus there exist such that . Since is arbitrary, it follows .

*Uniqueness:*

Suppose . Suppose . Thus and . Since is one-to-one, it follows . Thus there exists a unique element such that .

Thus .

**Soln13**

**(a)**

We have . It follows that . Clearly we have is compatible with , since . Thus it follows directly from Ex18 from sec-5.1, that is a function such that .

**(b)**

Since , it follows from ex16 from sec5.2 that is one-to-one.

Proof for is onto. Suppose . Since is onto, it follows that there exists an such that . Since and is an equivalence relation, it follows . Thus . Since is arbitrary, it follows that for every , there exists an element . Thus is onto.

**(c)**

Since is one-to-one and onto, it follows that exists. Suppose . Thus . It follows there exist such that and . Thus . Suppose . Thus , or . Since , it follows . Thus . Since , it follows . Since , it follows .

**(d)**

It seems like there is a little typo in the question, instead of , it seems it actually should be .

Suppose .

() Suppose . Suppose . Thus there must exist an element such that , or . Thus , which is equivalent to . Since , it follows , or . Thus from part(c) of this solution it follows that . Since , it follows . Since is arbitrary, it follows .

() Suppose . Suppose is an arbitrary element. Thus there must exist an element such that , or . Thus . By part(c) it follows that . Since , it follows . Thus . Since is arbitrary, it follows that .

**Soln14**

**(a)**

Suppose . Thus . It follows that there exist an element such that . Thus we have . Since is arbitrary, it follows that for all .

**(b)**

We know from Sec5.1 Ex-7, . Since , it follows . Also from part(a) of this solution we have . Thus is one-to-one and onto and .

**Soln15**

Suppose . Thus clearly . Now with , it follows from Soln14(b) that . Since , it follows .

**Soln16**

**(a)**

We have . Suppose . Thus we have . Thus . It is a quadratic equation, thus we have . Clearly . Thus .

**(b)**

Suppose such that where and .
Thus we have .
Thus .

Similarly . Thus .
Thus . Now from Soln14(b) we have .

**Soln17**

We shall prove by proving that and , where is set of all partial orders and is set of all strict partial order.

Proof of :

Suppose . Thus is strict partial order. Thus and are disjoint. It follows . Since and are disjoint, it follows . Thus .

Proof of :

Suppose . Thus is partial order. Thus . It follows . Since , it follows . Thus .

Thus is one-to-one and onto and .

**Soln18**

**(a)**

To prove is an equivalence relation, we need to prove that is reflexive, symmetric and transitive.

__Reflexive:__

Suppose . Suppose . Thus , or . It follows . Thus is reflexive.

__Symmetric:__

Suppose such that . Thus there exist an such that . It follows (by applying h on both sides). Similarly taking value corresponding to on both sides, . Thus . Thus is symmetric.

__Transitive:__

Suppose such that and . Thus for some , we have and . Thus we have . Thus . It follows is transitive.

Thus is an equivalence relation.

**(b)**

Suppose . Thus there exist an such that . Thus we have . Thus . Thus .

**(c)**

Suppose and . Thus there exist an such that . Thus . Applying on both sides, . Thus . Thus also has a fixed point = .