Can a regular map $\phi:\mathbb{A}^{m}_{k}\rightarrow\mathbb{A}^{n}_{k}$ always be extended to $\mathbb{P}^{m}_{k}\rightarrow\mathbb{P}^{n}_{k}$?
In particular, I am interested in the map \begin{eqnarray*} \phi: & \mathbb{A}^{7}_{k} & \longrightarrow & \mathbb{A}^{4}_{k}\\ & [x_{1},...,x_{7}] & \longmapsto & [x_{1},...,x_{4}] \end{eqnarray*}
I want to know whether or not that map is closed. If I could extend it to a map $\tilde{\phi}:\mathbb{P}^{7}\longmapsto\mathbb{P}^{4}$, this would indeed be so: Any closed subvariety $V$ of $\mathbb{A}^{7}$ is of the form $\tilde{V}\cap\mathbb{A}^{7}$ for some projective variety $\tilde{V}\subseteq\mathbb{P}^{7}$. The variety $\tilde{V}$ is projective, hence proper, so its image $\tilde{\phi}(\tilde{V})$ in $\mathbb{P}^{4}$ would be closed. Therefore $\phi(V)=\tilde{\phi}(\tilde{V})\cap\mathbb{A}^{4}$ would be closed as well. Or is there a mistake in this argument?
The only maps $\Bbb P^m\to \Bbb P^n$ with $m>n$ are the constant maps. See for instance here. As what you've written is not a constant map, it's not a map at all - for instance, the point $[0:\cdots:0:1]$ doesn't have an image under your "map".
There are easier ways to show your map $\Bbb A^7\to \Bbb A^4$ isn't closed. Consider the variety $V(x_1x_7-1)$: this is closed in $\Bbb A^7$, but it's image under the projection to $\Bbb A^4$ is $D(x_1)$, which is not closed.