Function defined in coursework has unmapped domain? Hi I was doing my Discrete Math coursework when I realise the function defined in the question has unmapped domain.

Note the definition of Zn*:



and in part c):



So basically you have f([3]) = [6] = [0], which maps to none of the codomain. How should I approach this?

Also, is the function injective? ( if both f( [ 7 ] ) = f( [ 1 ] ) = [ 2 ] ==> [ 7 ] =/= [ 1 ])??

Edit: I should add the domain and codomain do not seem to include [0].
You're correct that the definition fails to define a function $\mathbb Z^*_6\to\mathbb Z^*_6$.

Arguably that is a bug in the exercise. In that situation the best thing you can do is probably to point out the problem, and also explain that if we take the definition to define a function $\mathbb Z_6\to\mathbb Z_6$ instead, then it (is/is not) injective, and (is/is not) surjective.



Note that it is more common to take $\mathbb Z_n^\times$ to mean the group of all residue classes $[a]$ modulo $n$ where $a$ and $n$ are coprime. In that case, however, $\mathbb Z_6^\times$ would be $\{[1],[5]\}$ and the definition in the exercise still wouldn't define a function $\mathbb Z_6^\times \to \mathbb Z_6^\times$.
Okay, f([0])= [2][0]= (2)(0)= 0= [0], f([1])= [2][1]= 2= [2], f([3])= (2)(3)= 6= [0], f[4]= [2][4]=  8= [2], and f[5]= [2][5]= 10= [4].  The "codomain" is {[0], [2], [4]}.  I don't know what you mean by "f([3])= f([6])= [0] maps to none of the codomain".  You just said that it maps to [0]!  What is true is that there is no "x" in the domain that maps to [1] (or to [3] or [5]).  This function is not surjective or "onto" the range space {[0], [1], [2], [3], [4], [5]}.