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I think this function is injective, but had a tough time proving it. Here's my attempt.

Proof: Let $(x_1,x_2)$ be arbitrary elements of $\mathbb{P}_1$. Assume $g(x_1)=g(x_2)$, i.e. $\int_0^x x_1dt=\int_0^x x_2dt$. Integrating yields $\frac{1}{2}(x_1)^2=\frac{1}{2}(x_2)^2\implies \frac{1}{2}x^2=\frac{1}{2}x^2$. This implies that $x_1=x_2$, so the function $g$ is injective.

Brandon O'Neil
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  • Assume that $g(p_1)=g(p_2)$. Therefore, for all $x$, $\int_{0}^{x}p_1=\int_{0}^{x}p_2$. Since $p_1,p_2$ are continuous, those integrals are differentiable. Taking derivatives you get that $p_1(x)=p_2(x)$. – orole Feb 07 '18 at 16:54
  • If, as in your attempt, $x_1$ is an arbitrary element of $\mathbb P$, then integrating $x_1$ with respect to $t$ does not in general produce $\frac12(x_1)^2$. For example, if $x_1$ is $2t$, then the integral is $t^2$ (plus constant of integration), not $\frac12(2t)^2=2t^2$ (plus constant of integration). – Andreas Blass Feb 07 '18 at 18:06
  • A second error in your attempt is when you go from the indefinite integral to the definite integral. Since the integral is with respect to $t$, you should have substituted the limits of integration, $0$ and $x$, for $t$, not for $x_1$ and $x_2$. – Andreas Blass Feb 07 '18 at 18:08

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