Why is this curve $t \mapsto (t, t^2, t^3)$a smooth embedding?

Question

I have to show that the curve $f(t)= (t, t^2, t^3)$ embeds $\mathbb{R}$ into $\mathbb{R^3}$> And then find two independent functions that globally define the image.
My attempt:.
Here it is clearly an immersion since rank $df= 1$, map is one-one -this part too is clear to me. Now I have to show that it is a topological embedding to show that it is a smooth embedding.
The map is one one and onto its image, is continuous since each of the component function is. My question is: how do check if it's inverse is also continuous and also find its image.
And, I have got no clue how to go on about the solution of second part-finding two independent functions...
Any kind of hint/s are welcome.
Reference: Pollack, Differential Topology

Answer 1

You already have the injectivity and the immersion part. OK, now you can try to prove that the curve is a homeomorphism onto its image. But it is proved in Guillemin & Pollack's book that an equivalent condition (which I think in this case works faster) is that it is a proper map. Try to prove that the preimage through this curve of compact subsets of the image (which has the subspace topology inherited from $\mathbb{R^3}$) are compact in $\mathbb{R}$. This should not be hard for you, since the first coordinate is the identitity, hence taking $K\subset f(\mathbb{R})$ compact, its projection onto the $X$-axis must be compact too, but that's exactly $f^{-1}(K)$. So, it is proper, hence an embedding.
Now just observe that any $(x,y,z)\in f(\mathbb{R})$ must satisfy the pair of equations $y-x^2=z-x^3=0$. Check that his two functions are independent.