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I think this function is not injective, but I am unsure if I've correctly proven that.

Determine if the following statement is true or false and provide a proof to justify the answer. \begin{align*} g: \mathbb{R} & \to \mathbb{R} \\ g(x) &= \begin{cases} -x+1 & x>0\\ -x^2 & x \leq 0 \end{cases} \end{align*}

Here's my attempt. Proof: $g(-1)=-1=-1=g(2)$. $-1=-1$, but since $g(-1)=g(2)$, we can say that $g(-1)=g(2)\implies -1=2$ is false, and therefore for generic $x_1,x_2\in \mathbb{R}$, $g(x_1)=g(x_2)\implies x_1=x_2$ is false, so the function $g$ is not injective. $\blacksquare$

Am I missing something?

Brandon O'Neil
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    Your proof is correct, but you are overcomplicating the writing. You are showing that $g$ is not injective by way of the counterexample $g(-1) = -1 = g(2)$. That is really all you need to say! – Xander Henderson Feb 07 '18 at 21:27

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You have exposed two distinct points in the domain, $x=-1$ and $x=2$, with the same image under $g$. So $g$ is not injective. Done. A counterexample suffices.

Henno Brandsma
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That seems comprehensive to me: to prove that your function is not injective in its domain, you only need to show one example of $f(x)=f(y)$ while $x\not = y$; $x,y \in \mathbb R$.

ssalogel
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You're dealing with a claim:

Claim: $g$ is injective.

If the claim is true, then you need to provide a proof. Otherwise, you need to provide one counter-example, and you already did that. Therefore, the claim is false.

wjmolina
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