The hint for Question 4 may be helpful for Question 2, so I have put it first.
Question 4. This question asks you to find surjective functions and such that the function defined by is not surjective. One way to get a function that is not surjective is to take a constant function which takes only one value. For example, the function defined by for all . If you could arrange that and were surjective, and
for all , then you would be done.
Question 2. Your job is to define a function such that is neither injective nor surjective. By definition of surjective, is surjective if for each , there exists at least one such that . So to make sure that is not surjective, you need to make sure that there is some which is not equal to any for . Given this guide, try to define . If you find that with your definition, is also not injective, you have a suitable example.
Question 6. We haven’t covered relations yet in lectures. But you should be able to make an attempt at this and Question 7 by reading pages 35 and 36 of the lecture notes, issued in Monday’s lecture.
Here is a detailed solution to (6c). We define a relation on the set of people in a lecture room by if can see the eyes of . The relation is reflexive if for all we have , i.e. if everyone in the lecture room can see their own eyes. I think we can safely say this is false. (Not everyone can be looking in a mirror at once.) It is symmetric if
i.e., if and see the eyes of then can see the eyes of . This is true (at least if we make the kind assumption that everyone’s eyes are always open in a lecture). Finally, the relation is transitive if
In words, this says that if can see the eyes of , and can see the eyes of , then can see the eyes of . This is false: for example take to be the lecturer and and to be two people facing the lecturer. So is symmetric but not reflexive or transitive.
Question 7. Here are some examples for (a). We have since and
So you should draw an arrow from to . Also , since
so there should also be an arrow from to . Similarly there should be arrows from to , from to . There should also be a loop from to , since . In the diagram for Example 8.2, I used a single line with two arrow-heads instead of two separate arrows.
Question 8. For (c), you should use the formula for the inverse of a composition given at the top of page 34 of the lecture notes:
Applying this to we get
You should now use your formula from (b) for and then your formula from (a) for to turn the right-hand side of the equation above into an explicit formula for .