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How do you figue out whether this function is onto?

$\mathbb{Z}_3\rightarrow \mathbb{Z}_6:f(x)=2x$

Onto is of course is for all the element b in the codomain there exist an element a in the domain such that $f(a)=b$

Here the co domain is mod 6

So let $k\in\mathbb{Z}_6$

But I am not sure how to see if it is onto.

Fernando Martinez
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3 Answers3

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In general there exists an onto function with finite domain $A$ and finite codomain $B$ if and only if $ |A|\geq|B|$

hint: what number maps to $1,3$ or $5$ mod 6?

Asinomás
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  • I think odd numbers maps to 1,3,5 mod 6? – Fernando Martinez Jul 11 '14 at 18:53
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    No, 1 goes to 2, 2 goes to 4 and 3 goes to 6. – Asinomás Jul 11 '14 at 18:58
  • no I think I see now multiples of 3 – Fernando Martinez Jul 11 '14 at 18:58
  • I see so what you are saying is all the element in 3 mod 6 go to 6 mod 6 when multiplied by 2, for example 15 an element of 3 mod 6 goes to 6 mod (15)(2)=30, is in 6 mod 6. Maybe I am reading your theory wrong. – Fernando Martinez Jul 11 '14 at 19:07
  • No what I am saying is if $x$ leaves residue 1 when divided by $3$ then $2x$ leaves residue $2$ when divided by $6$, if $x$ leaves residue $2$ when divided by $3$ then $2x$ leaves residue $4$ when divided by $6$, and if $x$ is a multiple of $3$ then $2x$ is a multiple of $6$. – Asinomás Jul 11 '14 at 19:14
  • I see so there are classes for example 1 mod 6 $(-11,-5,1,7,13,19)$ which if we take mod 3 classes and muliply it by 2 will not equal this class. – Fernando Martinez Jul 11 '14 at 19:22
  • There may exist an onto function, but the one we are given may or may not be it. – Teepeemm Jul 11 '14 at 19:34
  • @Teepeemm really? could you provide an example? – Asinomás Jul 11 '14 at 19:35
  • Sorry. I should add that this statement can prove the nonexistence of an onto function, but it can't help us if an onto function exists. – Teepeemm Jul 11 '14 at 19:36
  • I think its not onto because there exist classes in the codomain mod 6 that will never be reached from mod 3 multiplied by 2 one example is 1 mod 6. And in onto all the codomian is subset of the range meaning they must be equal. Because for example all element is 0 mod 3 times 2 go to 1 mod 6, all element in 1 mod 3 times 2 are in 2 mod 6. Etc – Fernando Martinez Jul 11 '14 at 19:40
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    if you take a function with finite domain then the fibers of the image of the functions will form a partition of the domain, so the image of any function with finite domain has cardinality less than or equal to the domain. So if the codomain is larger than the finite domain, we can be sure the map is not onto. – Asinomás Jul 11 '14 at 19:41
  • To explain what I mean: there exists an onto function $f:A\to A$. But whether $f(a)=1$ is that onto function is a different matter. – Teepeemm Jul 11 '14 at 20:04
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Notice that if $x=3m+r$ then $2x=6m+2r$ where $r$ can only be $0,1,\text{or}\,2$.

user71352
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To put it a bit differently from Bananarama, the map gives you the values {$f(0),f(1),f(2)$}$(Mod 6)$. Even if these values are a different, can the map be onto?

user99680
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