No-one has asked for any hints on Sheet 9 yet, so here instead are some final thoughts on Sheet 8. Let be a function. This means that for every , there is an associated element , which we call the image of under . So if you are defining a function, it is your responsibility to ensure that is defined for each in the domain .
(Incidentally, some people confused this with the definition of injectivity, by writing something like ‘for each there exists some ‘. This doesn’t really make sense as it stands, because the function is never mentioned. If we change it to ‘for each there exists an element ‘ then it’s just the defining property of a function. Injectivity is something different: for each there exists at most one such that .)
I found it a little surprising how few people tried to use the method suggested in lectures to show that a function is injective. To repeat it one more time: suppose that and deduce, using whatever you know about that .
Please see the answers on Moodle for examples of the approved method, especially for Question 5(a). Of course any correct proof is acceptable, and there were several people who gave correct, if sometimes a little long-winded, proofs of 5(b) using proof by contradiction.
Question 2 asked for a function that was neither injective nor surjective. Several people tried something like . This is not a bad attempt, but there is one problem: we have , and is not in the codomain of . You could get round this by taking . However there is no need to specify by a quadratic expression in this way: when the domain is a finite set it is usually easier just to list the images. So you could just say: let , and .