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I am little unclear on this problem. I need to prove that this is a one-one function. I know that I need to show $f(x) = f(y) \Rightarrow x = y$. But I don't know how this applies to piecewise functions.

$f:x \rightarrow \begin {cases} |x| - 1 & x \leq -1 \\ -x^2 & x \geq 0 \end {cases}$

Thanks for your help.

Edit: The answer by @Joe seems to make sense to me. But can you guys clarify this. Piecewise functions generally graph out as fragments. Does the one-one function property $f(x) = f(y) \Rightarrow x = y$ differ in that case?

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    In answer to the edit, the one-to-one property doesn't differ in definition. A better way to think might be if $x$ and $y$ are two different numbers, then $f(x)$ and $f(y)$ are two different numbers. In terms of graphs, you shouldn't be able to draw a horizontal line that intersects the graph in more than one place.2011-06-27

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This function is not one-to-one, since $f(0)=f(-1)=0$.

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    I delete my answer since the question has been edited2011-06-27
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The definition of a one-to-one function (also called injective sometimes) is that two different inputs are mapped to two different outputs.

That is, if $f(x)=f(y)$ then $x=y$. Whether or not the function is defined by a formula, piecewise, not continuous, or polynomial... it does not matter.