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Let $h,g$ be not injective functions, can function $f:\mathbb{R}\rightarrow \mathbb{R}^2$ such that $f(x) = (h(x), g(x))$ be injective?

I know that, if I pick polynomials for $h$ and $g$, then may be not injective, if picked carefully. For example, I can check zeros of the polynomials, whether they collide or not, but I am really not sure whether there exist other, nontrivial counterexample to disprove injectivity of $f$.

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$f$ can be injective. For instance, $g(x) = x^2, g(x) = (x + 1)^2$. Then $f(x) = (x^2, (x+1)^2)$ is injective, since $g(a) = g(b)$ and $a \neq b$ means $a = -b$, but $h(a) \neq h(-a)$. Thus, if $a$ and $b$ are such that the first component of $f(a)$ is equal to the first component of $f(b)$, then the second components are different.

As for a non-trivial, non-injective $f$, one may take $g(x) = x^2$ and $h(x) = x^3 - x$. Then $f(-1) = f(1) = (1, 0)$.

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  • $\begingroup$ Yes, that's what I thought. If the given polynomials share some roots where x is different, they become non-injective. $\endgroup$ – PeterBocan Nov 22 '16 at 23:00
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    $\begingroup$ @PeterBocan It doesn't have to be roots, though. Take $f(x) = (x^3 - x + 5, x^2 + 5$, for which $f(-1) = f(1) = (5, 6)$. $\endgroup$ – Arthur Nov 22 '16 at 23:06
  • $\begingroup$ Yep, I know, I am just thinking about how to find/not find such non-trivial counterexamples to injectivity in general terms for any two polynomials. You know, generally speaking, of course. Otherwise you would have to evaluate all numbers... $\endgroup$ – PeterBocan Nov 22 '16 at 23:16
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Let $h(x)=x^2$, and $g(x)=(x+1)^2$. $h$ and $g$ are not injective.

But,if $f(x)=f(y)$, we have $x^2=y^2$ and $(x+1)^2=(y+1)^2$.

So, $(x+1)^2-x^2-1=(y+1)^2-y^2-1$.

And then $2x=2y$. So $x=y$.

So, $f$ is injective.

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