During a lecture in algebraic geometry the following fact was mentioned without proof
If we consider an affine variety $X\subset \mathbb{A}^n$ and the map $f_n:\mathbb{A}^n\to \mathbb{A}^{n-1}$ $f(x_1,\dots,x_n)=(x_1,\dots, x_{n-1})$ then if $(0:0:\dots:1)$ is not in the projective closure of X then $f(X)$ is closed in $\mathbb{A}^{n-1}$.
The only suggestion was to look at the (to be proven to be) commutative diagram (where $\varphi_n(x_1,\dots x_n)=(1:x_1,\dots, x_n)$ and $\overline{f_n}(x_0:\dots:x_n)=(x_0:\dots :x_{n-1}$))
$$ \begin{matrix} X& \rightarrow_{\varphi_n}& \overline{X}\\ \downarrow_{f_n} & & \downarrow_{\overline{f}_n}\\ \mathbb{A}^{n-1} &\rightarrow_{\varphi_{n-1}}& \mathbb{P}^{n-1} \end{matrix}$$
However I don't see how this makes thing easier, it looks to me that removing the last coordinate from an affine variety is as hard as doing it from a projective one (id est quite hard given that all my attempts failed), on the other hand proving this is a commutative diagram is quite easy.
Am I missing something obvious?
This looks a lot like a homework or an exercise you have to solve, so I don't think is very usueful to spoon feed you the solution.
What I can suggest is to focus on the relationship between the two vertical functions and aftert you got that look at what does $\overline{f}_n$ to closed sets.