Map from $\mathbb{A}^1 \rightarrow \mathbb{A}^2$ Let the map $\varphi_n:\mathbb{A}^1 \rightarrow \mathbb{A}^2$ be defined by $t\rightarrow(t^2,t^n)$. 
-Show that if n is even, the image of $\varphi_n$ is isomorphic to $\mathbb{A}^1$ and $\varphi_n$ is 2:1 away from 0.
-Show that if n is odd, $\varphi_n$ is bijective, and give a rational inverse of it.
For the even case: I believe that I've shown that $\varphi_n$ is exactly 2:1, and I believe that I've shown that the standard curve given by $y=x^{\frac n2}$ is the image of $\varphi_n$. How do I go about showing that this is isomorphic to $\mathbb{A}^1$?
For the odd case: I'm not really sure where to start here, I used the same process that I did for the even case, and I think the function $t=\frac{y}{x^m}$ where $(x,y)=(t^2,t^n)=(t^2,t^{2m+1})$ for some m. How to show that this is bijective?
Thanks for any help you guys can offer!

For the even case, you are showing that the image is isomorphic to $\mathbb{A}^1$. It is not necessary (and not true) that $\varphi_n$ is the isomorphism. In fact, your answer almost contains the map from $\mathbb{A}^1$ to $\varphi_n(\mathbb{A}^1)$ and the map in the other direction. (You will need to check that they are mutual inverses of course).
In the odd case, I think you want to show that it is bijective onto its image ($\varphi_n$ is not surjective onto $\mathbb A^2$ because $|y|$ is determined by $|x|$). Pick a point $(x,y)$ in the image of $\varphi_n$. Using what you said in your question, you need to show that $\frac{y}{x^m} = \frac{t^{2m+1}}{(t^2)^m}$ gives you an inverse to the map $\varphi_n$. 
