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Is f(m,n) = absolute value of (n) an onto function? How would you prove it? It only depends on n so m could be any (integer). Also, this function is from the set of Integer coordinates to an integer: Z x Z -> Z

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No. How about $-1$? It is not in the range. Your function is projection onto the first coordinate composed with absolute value. Projection is onto, but the absolute value still is not.

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    I was thinking that too about the negative numbers, but for onto, don't you have to just prove that for all numbers in the Range, there will be something in the domain that maps onto them? – 2017-02-02
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    And since we end up with only positive numbers because of absolute value, isn't it still ok because they get mapped to? Or is it automatically not onto because the range doesn't include the set of {all integers} – 2017-02-02
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    @Alex If the function were onto, that is how one would prove it. However, the function is not onto, as I have demonstrated by finding something in the range ($-1$) whose has nothing in the domain which maps to it under the function. – 2017-02-02