[Math] One-One function check with piecewise function

Question

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?

Answer 1

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