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Suppose that $A$ is a mapping from a set $S$ to itself and $A(A(x)) = x$ for all $x \in S.$ Prove that $A$ is one-to-one and onto.

Can someone please break this down, define one to one, and onto? I am new to this terminology.

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    $\begingroup$ Why not use a search engine to find definitions for those terms? $\endgroup$ – Trevor Wilson Oct 24 '13 at 4:00
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    $\begingroup$ To become a better mathematician, I figured it best to find help from better mathematicians. I appreciate your recommendation. $\endgroup$ – Aspiring Mathematician Oct 24 '13 at 4:08
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    $\begingroup$ Probably some people here will be happy to help you in any case, but looking up the terms yourself has (at least) two advantages: (1) it saves answerers the trouble of duplicating effort, and (2) it allows you to ask a more focused question, or perhaps to answer the question yourself. $\endgroup$ – Trevor Wilson Oct 24 '13 at 4:22
  • $\begingroup$ Also, people might misinterpret the question as a request to solve the exercise, thereby depriving you of the opportunity to solve it yourself. $\endgroup$ – Trevor Wilson Oct 24 '13 at 4:23
  • $\begingroup$ @AspiringMathematician Please, try to make the title of your questions more informative. E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. For more information on choosing a good title, see this post. $\endgroup$ – Lord_Farin Oct 24 '13 at 17:33
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A function $f$ is called $1-1$ if $f(x) = f(y) \implies x = y$.

So suppose that

$$A(x) = A(y)$$

Applying $A$ to both sides, we see that

$$A(A(x)) = A(A(y))$$

But how can you write $A(A(x))$ and $A(A(y))$?


A function $g$ is called onto if for every $y$, there exists an $x$ for which $f(x) = y$. Given $y$, write

$$y = A(A(y))$$

So what input to $A$ gives $y$ as an output?

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    $\begingroup$ From this is it correct that y as an input to A gives y as an output? $\endgroup$ – Aspiring Mathematician Oct 24 '13 at 3:50
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    $\begingroup$ @PythagorasWannaBe Not quite. $A(y)$ as an input gives $y$ as an output. $\endgroup$ – user61527 Oct 24 '13 at 3:51
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    $\begingroup$ I see, I was missing the notation A(y) = y $\endgroup$ – Aspiring Mathematician Oct 24 '13 at 3:53
  • $\begingroup$ @PythagorasWannaBe No, $A(y) = y$ is not true. $\endgroup$ – user61527 Oct 24 '13 at 3:54
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    $\begingroup$ Then what input to A gives y as an output? $\endgroup$ – Aspiring Mathematician Oct 24 '13 at 3:57

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