Showing that $\mathbb{R}^2 /N \overset{\simeq}\to g_0$ where $g_0$ denotes a line 
Consider in the $\mathbb{R}$-Vectorspace $\mathbb{R}^2$ the sub vectorspace $N:= \mathbb{R}w$ where $w\neq 0$. For $w_0 \neq 0$ let $g_0:= \mathbb{R}w_0 \subset \mathbb{R}^2$ be the sub vectorspace (namely a line) such that $g_0 \cap N= \lbrace 0 \rbrace$.
Show that the mapping $$\psi : \begin{cases} \mathbb{R}^2/N & \longrightarrow g_0 \\ x+N & \longmapsto (x+N)\cap g_0 \end{cases} $$
is an isomorphism.

Note: $x+N=[x]_N = \lbrace x' \in \mathbb{R}^2 \mid \exists n \in N: x'=x+n\rbrace$
My approach: After a lot work I already managed to show that $\forall x \in \mathbb{R}^2: |(x+N) \cap g_0| = 1$ which I think is a big step towards the solution to this problem. Geometrically this is of course trivial. On paper this requires a lot of steps therefore I am not going to repeat them here.
My first issue starts with the attempt to show that $\psi$ is a homomorphism. This isn't even geometrically trivial (to me) but after drawing it several times on paper the result is evident to me. That is at least for arbitrary $[x]_n, [y]_n \in \mathbb{R}^2 / N$ I want to show that $$\psi([x]_N + [y]_N)=([x]_N + [y]_N)\cap g_0 \overset{!}{=}[x]_N \cap g_0 + [y]_N \cap g_0 = \psi([x]_N) + \psi([y]_N) $$
But I just couldn't figure out yet how to do that. I know that the intersection will yield a point, but it is not clear that the point given by $$([x]_N + [y]_N)\cap g_0 = ([x+y]_N)\cap g_0 $$
is the same than the point given by $$ [x]_N \cap g_0 + [y]_N \cap g_0 $$

For Bijectivity: Since $\forall x \in \mathbb{R}^2$ it is true that $|(x+N) \cap g_0|=1$ the surjectivity should follow from that. For injectivity we have $g_0 \cap N = \lbrace 0 \rbrace$ which means that for arbitrary $[x]_N \in \ker \psi$ with $\psi ([x]_N) = \lbrace 0 \rbrace$ it must follow that $[x]_N = [0]_N$. This feels vaguely formulated to me, but I believe it goes in the right direction.
 A: Use that $\{w,\,w_0\}$ is a basis of $\mathbb{R^2}$. Now everything relies on the following fact (prove it!):

The given map can be written as
  $$\psi : \begin{cases} \mathbb{R}^2/N & \longrightarrow g_0 \\ x+N & \longmapsto \beta\,w_0 \quad \text{where}\quad x=\alpha w+\beta w_0\end{cases}$$

Now

$\psi$ is well defined

Let $x,y$ be such that $x+N=y+N$. Then $y-x\in N$ and we can write $y-x=\lambda w$ for some $\lambda\in\mathbb R$. In particular, if
$$x=\alpha w+\beta w_0 \ \Longrightarrow\ y = x+(y-x) = (\alpha+\lambda) w+\beta w_0$$
And since $\psi$ depends only on the second coordinate of the representative, we have
$$\psi(x + N) = \beta w_0 = \psi(y + N) $$

$\psi$ is a homomorphism.

Take $x,y\in \mathbb R^2$, then:
$$
\begin{cases} x=\alpha w+\beta w_0 \\ y=\gamma\, w+\delta\, w_0 \end{cases}
\quad \Longrightarrow \quad x+y = (\alpha+\gamma)w + (\beta+\delta)w_0
$$
Now it is trivial to check that
$$\psi(x+y+N) = (\beta+\delta)w_0 = \beta w_0 + \delta w_0 = \psi(x+N) + \psi(y+N)$$
In the same vein you can check that
$$\psi(\lambda x+N) = \lambda \psi(x+N)$$

$\psi$ is bijective

For any $\beta w_0\in g_0$ you can check that $\beta w_0 + N\in\mathbb{R}^2/N$ is the only element such that $\psi(\beta w_0 + N) = \beta w_0$. You can fill in the details.
