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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.

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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?

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  • $\begingroup$ I think odd numbers maps to 1,3,5 mod 6? $\endgroup$ – Fernando Martinez Jul 11 '14 at 18:53
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    $\begingroup$ No, 1 goes to 2, 2 goes to 4 and 3 goes to 6. $\endgroup$ – Jorge Jul 11 '14 at 18:58
  • $\begingroup$ no I think I see now multiples of 3 $\endgroup$ – Fernando Martinez Jul 11 '14 at 18:58
  • $\begingroup$ 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. $\endgroup$ – Fernando Martinez Jul 11 '14 at 19:07
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    $\begingroup$ 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. $\endgroup$ – Jorge Jul 11 '14 at 19:41
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Notice that if $x=3m+r$ then $2x=6m+2r$ where $r$ can only be $0,1,\text{or}\,2$.

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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?

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