e) Working only from the definitions of function, surjectivity, and injectivity, and not from other prior results, prove your answer for either line c) or line d) above. If there is an entry in the middle column, pin down the conditions under which that condition will hold of f.

No answers on this one, but a hint for part e): think of the function as a relation. That’s often helpful in working with functions between finite sets.

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Reflexivity is an existence property; a possession property. If you contain this entire particular set of pairs, you’re reflexive. If not, then not.

Symmetry and transitivity are implications; closure properties. Now some pairs don’t come for free – they require other pairs be added as well. Suppose you’re having a party. You have two friends who just started dating and are very attached – so if you have one of them over to the party you also have to have the other. That’s symmetry. You have another set of friends, a married couple with a new baby, who can’t get a babysitter. You can have just the husband or just the wife, but if you want to have both of them come, the baby has to come too. That’s transitivity.

In a numerical example: let A = {1,2,3,4,5}. Start building a relation R. Put (1,2) into R. If we want a symmetric R, we must also add (2,1). {(1,2)} is transitive, so for that we don’t need to add anything. However, if we put in (2,1) we must put in (1,1) and (2,2) for R to be transitive; this is called closing R under transitivity.

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Answer after the jump.

To succeed in this requires thinking of functions in the sense of “relations such that each element appears exactly once as the first element of a pair” and ignoring any pull toward functions as equations. The most general answer is to partition A, fix one element of each partition set, and map every element of that partition set to the fixed element. For example, map 1 and 2 both to 1, and map 3, 4, and 5 all to 4. Any constant function from A to A satisfies the requirements, at the size-1 partition end; the identity function arises from a partition of size 5. These are transitive because the only pairs of the form (a,b), (b,c) in the relation are those where b=c.

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