Re: Functions as relations.

we don't find "a" pre-image of f(3) (which is what the 3 with the little bar on top is), we find the entire set of pre-images.

the elements of Z6 aren't "integers" they are "equivalence classes of integers". for example:

[0] = multiples of 6 = {0+6k: k in Z} = {....-12,-6,0,6,12,18,....}

[1] = multiples of 6 + 1 = {1+6k : k in Z} = {....-11,-5,1,7,13,19,...}

and so on.

note that [0] = [6] = [12], etc.

[1] = [7] = [13], etc.

in fact, [k] = [m], if and only if k - m is a multiple of 6, that is, if and only if k ≡ m (mod 6). the elements [k] are often called congruence classes, or residue classes modulo 6.

the canonical map f:Z-->Z6 is k-->[k], or written another way: k-->k+6Z.

so if f(k) = f(3) (that is, k is in f^-1([3])), it must be the case that f(k) = 3+6Z, that is, k is of the form: k = 3+6m for some integer m.

Re: Functions as relations.

so for a pre image of f(3) is {...-15,-9,-3,3,9,15,21...}?

Re: Functions as relations.

indeed, there is a 1-1 correspondence between the cosets of 6Z, k+6Z, and the equivalence classes [k] of Z6. this is no accident, since if you look close enough, you see that they are just different ways of naming "the same thing". there is even a name for this correspondence: it's called the First Isomorphism Theorem, and it crops up in many different places under many different names:

(partition by image elements, in sets and functions

FIT in group theory, ring theory, module theory, and algebras

rank-nullity theorem in vector spaces

splitting lemma for short exact sequences...don't worry about this if you don't know what all of these are).

Re: Functions as relations.

Thank you, i totally understand another explanation by you. (Rofl)