The elements of $\mathbb{Z}^2$ would all be positive perfect squares, so when mapped onto integers, it is injective right?
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$1 \mapsto 1, 4 \mapsto 1, 9 \mapsto 1, 16 \mapsto 1, \dots$ – Jun 16 '14 at 14:50
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how can 4 map to 1? – baba Jun 16 '14 at 14:51
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5$Z^2$ usually means the set of ordered pairs of integers, not the set of squares. – MJD Jun 16 '14 at 14:51
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@baba Why not? I just thought of the first function that comes to mind, which is constant. – Jun 16 '14 at 14:51
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so all functions from Z^2 to Z are therefore NOT injective? – baba Jun 16 '14 at 14:52
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3Some are, some aren't. – MJD Jun 16 '14 at 14:53
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2I do not understand the question as it is stated.. are you asking "does there exists " a function $f : \mathbb{Z}^2\rightarrow \mathbb{Z}$ which is injective which is not constant function?? – Jun 16 '14 at 14:53
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In the first comment, @user61527 shows a counterexample (assuming that what you meant to ask was "are all functions ... injective?"). – barak manos Jun 16 '14 at 14:54
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yes, I was wondering if all functions were injective for the given mapping. Thank you for the responses – baba Jun 16 '14 at 15:04
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1A function is a mapping. It does not make any sense to ask "all functions are injective for a given mapping". – MJD Jun 16 '14 at 15:09
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I meant for the given domain and codomain. Sorry – baba Jun 16 '14 at 15:19
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Usually $Z^2$ denotes the cartesian product of the integers with the integers. For a proof there is a bijection between them see here.
We now discuss the case if $Z^2 $ denotes the set of perfect squares:
The function that sends every integer $k$ to $k^2$ is not injective, since $1^2=(-1)^2$. If you restrict it to non-negative elements then that function is indeed an injection.
There is a bijection between them though, just map the positive integers to the even squares and the negative integers to the odd squares. So $f(0)=0$ and $f(n)=(2n)^2$ if $n$ is positive and $f(n)=(-2n-1)^2$ if $n$ is negative.
