If we have two functions, $f:\mathbb{Z}\rightarrow\mathbb{Z}$ and $g:\mathbb{Z}\rightarrow\mathbb{Z}$ where $g(f(a))=a$ for every integer $a$, how do we satisfy these conditions so that $f$ and $g$ are not bijective?
Form a compositional identity function
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elementary-number-theory
functions
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0The question is not really clear. Which conditions are you referring to? – 2011-11-17
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0It should not be that hard to find such an example. Similar example was given for N instead of Z in another your question: http://math.stackexchange.com/questions/82961/why-do-these-functions-have-to-be-bijective/82964#82964 – 2011-11-17
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0I apologize for my ambiguity. What I mean to ask is: can we find functions $f$ and $g$ in which neither is bijective so that $f:\mathbb{Z}\rightarrow\mathbb{Z}$ and $g:\mathbb{Z}\rightarrow\mathbb{Z}$ where $g(f(a))=a$ for every integer $a$ – 2011-11-17
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0@MartinSleziak: it seems much tougher to find examples for $\mathbb{Z}$ because of the difficulty of dealing with negative numbers. So far I have come up empty – 2011-11-17
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0What about taking the maps from that answer and extend them by putting $f(a)=g(a)=a$ for each $a\le 0$? – 2011-11-17
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0But if $f(a)=g(a)=a$ then we would have two bijective functions, which is what I am trying to avoid – 2011-11-18
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0@johnnymath: Martin Sleziak's point was to use the function on $\mathbb{N}$ from Alon Amit's post, then extend it to negatives with the identity. The section on $\mathbb{N}$ has already violated bijectivity. – 2011-11-18