## Chapter - 5, Functions

### Section - 5.4 - Images and Inverse Images: A Research Project

### Summary

- Suppose and . Then the image of under is the set defined as follows:

. - If , then the inverse image of under is the set defined as follows:

. - In this section, we also learned a way to come up with counter examples. It might not help always for finding counter examples but it may
atleast give a starting point. The outline is as follows:
- Try to prove the theorem.
- If theorem is not correct, we will get stuck at some point.
- Try to use the information gathered from the point where we are stuck in the proof to come up with counter example.

**Soln1**

**(a)**

Yes it will always be true. Proof:

()Suppose . Thus for some , . Thus we have two cases:

Case 1:

Thus and , it follows . Or we can also say .

Case 2:

Thus and , it follows . Or we can also say .

Thus from both cases, . Since is arbitrary, it follows that .

() Suppose . Thus we have two possible cases:

Case 1: .

Thus for some , . Since , it follows . Since and ,
it follows .

Case 2: .

Thus for some , . Since , it follows . Since and ,
it follows .

Thus from both cases . Since is arbitrary, it follows that .

Since and , it follows that .

**(b)**

No, it is not true. Counter example:

Suppose .

.

Thus, .

.

.

Also, , Thus .

Clearly .

**(c)**

No, it is not true. Counter example:

Note: Earlier solution was wrong. Check the comments.

Updated Solution:

Suppose .

.

Clearly .

Now, and . Clearly but .

Old(Wrong) Solution:

Suppose .

.

Clearly .

Now, and . Clearly . Thus but .

**Soln2**

**(a)**

Yes, it will be always true. Proof:

()Suppose . Thus and . Since and , it follows . Similarly, since and , it follows . Thus . Since is arbitrary, it follows .

()Suppose . Thus and . It follows that and and . It follows . Since and , it follows that . Since is arbitrary, it follows .

Thus from both directions, we can conclude

**(b)**

Yes, it will be always true. Proof:

()Suppose . Thus and f(x) \in Y \lor f(x) \in Z $$. Thus we have two cases:

Case 1:

Since and , it follows .

Case 2:

Since and , it follows .

Thus we have or . It follows .

Since is arbitrary, it follows .

()Suppose . Thus or . It follows we have following cases:

Case 1:

Thus and . Or we can also say .

Case 2:

Thus and . Or we can also say .

It follows from all possible cases that and . Thus . Since is arbitrary, it follows that .

Thus we have and . We can conclude that .

**(c)**

Yes, it will be always true. Proof:

()Suppose . Thus and . Thus . Since and , it follows . Also, since and , it follows . Since and , it follows that . Since is arbitrary, it follows .

()Suppose . Thus and . Since , it follows and . Since and , it follows . Thus we have and . It follows . Since and , it follows that . Since is arbitrary, it follows that .

Now we can conclude from both directions that .

**(d)**

No, it is not always true. Counter example:

Suppose .

Thus .

Also, and . Thus .

Thus even if , .

**Soln3** False.

Suppose .

Thus and . Thus . Thus if is not one-to-one, theorem is not correct.

**Soln4** False.

Suppose .

Also, and . Thus . Thus if is not onto, theorem is incorrect.

**Soln5** TODO

**Soln6**

Suppose denotes inverse image of under . Thus .

Also suppose denotes image of under . Thus .

()Suppose . Thus and . Suppose . Thus and . Thus there exists an element such that . It follows that . Thus .

()Suppose . Thus for some element we have . Since is one-to-one and onto, and , it follows . Since , it follows . Since and , it follows . Thus .

Since and , we can conclude that .