Injective curve

Question

How can i show that the curve $\gamma:\mathbb{R}\to\mathbb{R}^2$ defined by \begin{equation} \gamma(t)=\left(\frac{t}{1+t^2},\frac{t}{1+t^4}\right) \end{equation} is injective (without using algebraic geometry techniques)? Does a general way to show it exist for general curves?

Answer 1

$$ f:t \mapsto \frac{t}{1+t^2} $$ is an "almost injective" function: $f(x)=f(y)$ iff $xy=1$. In such a case, however, $$ g(x)=\frac{x}{1+x^4}\neq \frac{y}{1+y^4} = g(y)$$ unless $x=y=1$.

Answer 2

There is in fact a standard method for such problems, relying on the definition of injectivity. Here it is: let us assume that

$$\gamma(s)=\gamma(t)$$

with the objective to deduce that necessarily $s=t$.

or, in an equivalent way, that there doesn't exist a couple $(s,t)$ such that

$$\gamma(s)=\gamma(t) \ \ \ \text{with} \ s \neq t \ \ \ \ (1)$$

Assume that such a couple exist: (1) is equivalent to $$x(s)-x(t)=0 \ \text{and} \ \ y(s)-y(t)=0$$

Easy computations where we cautiously preserve factors $(s-t)$ bring the equivalent conditions:

$$(s-t)(st-1)=0 \ \text{and} \ (s-t) (s^2t^2+st^3+s^3t-1)$$

But, by hypothesis $s \neq t$; therefore, the preceding double condition is equivalent to:

$$st=1 \ \text{and} \ s^2t^2+st^3+s^3t-1=0 \ \ \ \ (2)$$

Using relationship $st=1$ in the second equation of (2) gives $s^2+t^2=0$ which is not compatible with $st=1$. Therefore we have obtained a contradiction.

Remark: one can observe that the curve of $\gamma$ has the origin as a limit point when $t \rightarrow \infty$.