Questions about a polynomial map I want to solve the following problem.

Let $\mathcal{C}=\mathrm{Im}\varphi$, where $\varphi:\mathbb{C}\to\mathbb{C}^{2}$ is the polynomial map given by $$\varphi(t)=(t^{4}-2t^{2}+2,t^{4}-t^{2}-1)$$
  
  
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*Prove that $\varphi:\mathbb{C}\to\mathbb{C}^{2}$ is not dominant and $\varphi:\mathbb{C}\to\mathcal{C}$ is dominant but not bijective.
  
*Modifying $t$ and $\varphi$, prove that $\mathcal{C}$ is polynomically isomorphic to a straight line.
  
*As a consequence, identify $\mathcal{C}$.
  

I know the following facts, but I have questions about them too:

Let $\varphi: V \to W$ be a polynomial map and $D_V (\varphi)$ be the domain of $\varphi$. Then we say that $\varphi$ is dominant if $\varphi(D_V(\varphi))$ is dense in $W$.

Is $\varphi(D_V(\varphi))$ the same as $\operatorname{Im}(\varphi)$? I suspect that it is not because of this use of notation, but if you take the image of the domain, it is the image set, right?

A polynomial map $\varphi: V\to W$ defines a homomorphism $\varphi ^*: \mathbb C[W] \to \mathbb C[V]$ given by $\varphi^*(g)=g\varphi$.

What is $g$? I think it must be a polynomial function $g:W\to \mathbb C$. But I don't know how to find $\varphi^*$ in my problem (I don't know even if it is unique), which would be useful because I could use the following:

$\varphi$ is dominant if and only if $\varphi^*$ is injective.

I think this is the way to solve the first part of the exercise.
I don't know what the second part is asking (what is to modify $t$ and $\varphi$ and how it helps to show the isomorphism).
The third part must be straightforward when I get the first two, I hope.
 A: This is quite an instuctive example. The map $\varphi^\ast$ in your case is the following map:
\begin{align*}
\varphi^\ast:\Bbb C[x,y] &\longrightarrow \Bbb C[t] \\
x &\longmapsto t^4 -2t^2 + 2 \\
y &\longmapsto t^4 - t^2 - 1
\end{align*}
Indeed, if $g\in\Bbb C[x,y]$ is any polynomial function on $W=\Bbb C^2$, then you have
$$\varphi^\ast(g)(t)=g(t^4-2t^2+2,t^4-t^2-1)=g(\varphi(t))$$
which means $\varphi^\ast(g)=g\circ \varphi$.
Now for the next bit I cheated, I do not know what the exercise really wants you to do, but using this singular code
ring R=0,(t),dp;
ring S=0,(x,y),dp;
setring R;
map f=S,ideal(t^4 -2*t^2 + 2, t^4 - t^2 - 1);
setring S;
ideal ker=kernel(R,f)
ker

you get that $\varphi^\ast(f) = 0$ for 
$$
f = (x-y)^2 -5x+4y+5 = x^2 - 2xy + y^2 - 5x + 4y + 5
$$
In particular, $\varphi$ is not dominant because $\varphi^\ast$ is not injective. But why is that? It's because you have found the function $f$ on $W$ which is not the zero function and it vanishes on all of $\mathcal C$. If $\mathcal C$ was dense, this would be impossible because the vanishing of a polynomial is always very very tiny.
It is quite obvious that any map corestricted to its image is dominant, and $\varphi$ is also clearly not bijective because $t$ and $-t$ will always map to the same point. 
So, by "modifying" $\varphi$, I would suppose they suggest to compose it with linear automorphisms of $\Bbb C^2$, yielding an image that is isomorphic to $\mathcal C$. One interesting matrix to apply would be 
$$
A=\begin{pmatrix}
-1 & 2 \\
-1 & 1
\end{pmatrix}
$$
which yields
$A(\varphi(t)) = (t^4-4,t^2-3)$ 
and then substitute $s:=t^2$ to get $A(\varphi(s))=(s^2-4,s-3)$. The second component here gives you your isomorphism to $\Bbb C$.
Here is some more context. Some of the results I use might not be available to you, but it might be interesting for the big picture. I claim that $\mathcal C \ne \bar{\mathcal{C}} = Z(f)$, so the image of the polynomial map $\varphi$ is not closed in $\Bbb C^2$. Indeed, the inclusion $\bar{\mathcal C}\subseteq Z(f)$ is clear because $f$ vanishes on $\mathcal C$. Furthermore, $\dim(\bar{\mathcal C})\ge\dim(\mathcal C)=1$. Finally, $\mathcal C$ is irreducible as the image under a continuous map, of the irreducible variety $\Bbb C^2$. This means that $\bar{\mathcal C}$ is irreducible and closed of dimension one and contained in $Z(f)$. If we can show that $f$ is irreducible, they must be equal. It is easy to check that $\partial_x f + \partial_y f = -1$, so $Z(f)$ is smooth and in particular irreducible. Therefore, $\bar{\mathcal C}$ is not isomorphic to the affine line, it is an irreducible quadratic curve. Hence, $\mathcal C \ne \bar{\mathcal C}$, i.e. the image of the polynomial map $\varphi$ is not closed.
