This question is closely related to this one. Again consider the group $G := PSL(2, \mathbb F)$ over some field $\mathbb F$.
Then as written in the other post, there are two natural actions associated with $G$, one is by considering the points $\mathbb F^2$ and look at the induced action of $SL(2,\mathbb F)$ (as $PSL(2,\mathbb F)$ is just a quotient of $SL(2, \mathbb F)$), another by viewing $G$ as consisting of "Möbius"-like maps of the form $$ x \mapsto \frac{ax + bc}{cx + d} $$ where $ad - bc = 1$ (remark: you get the same set of mappings by just asserting that $ad - bc$ must be a square) and let them act on $\mathbb F_{\infty} = \mathbb F \cup \{\infty\}$ (the extended field, or the projective line). Now I want to show that both actions are equivalent, i.e. we have essentially the same actions by identifiying the points $\mathbb F^2$ with the projective line $\mathbb F_{\infty}$.
I started to give such an indentification, and then started to compute if the actions are equivalent, I give all these computations. In case iv) and v) I got stuck. So I am asking if anybody could help me resolving these points?
Now the computations: Given two actions of $G := PSL(2, \mathbb F)$ on two sets, $\Omega := \mathbb F^2$ and $\Gamma := \mathbb F_{\infty}$. Then an operation of a group $G$ on two sets $\Omega, \Gamma$ is called equivalent if there exists a bijection $\lambda : \Omega \to \Gamma$ such that $$ \lambda(\alpha^g) = \lambda(\alpha)^g $$ for all $g \in G$. The bijection is $\lambda((x,y)^T) = x/y$ if $y \ne 1$ and $\lambda((x,0)^T) := \infty$ otherwise. An element of $g \in PSL(2,\mathbb F)$ is represented by four numbers $a,b,c,d \in \mathbb F$ such that $ad - bc = 1$. This could be viewed as a matrix modulo $\{I, -I\}$, or as a mapping as written at the beginning. So if $(x,y)^T \in \mathbb F^2$ we define the action by matrix multiplication, and for $x \in \mathbb F_{\infty}$ by applying the mapping. Now let $g \in PSL(2, \mathbb F)$ and let $a,b,c,d$ the corresponding numbers (up to a common constant).
i) $y \ne 0, cx + dy \ne 0$
Then \begin{align*} \lambda\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \right) & = \lambda\left( \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix} \right) = \frac{ax + by}{cx + dy} = \frac{a(x/y) + b}{c(x/y) + d} = \lambda((x,y)^T)^g. \end{align*}
ii) $y = 0, cx + dy \ne 0$. Then $x \ne 0$ and $c \ne 0$ is implied. \begin{align*} \lambda\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \right) & = \lambda\left( \begin{pmatrix} ax \\ cx \end{pmatrix} \right) = \frac{ax}{cx} = \frac{a}{c} = \infty^g = \lambda((x,y)^T)^g. \end{align*}
iii) $y = 0, cx + dy = 0, c = 0$. \begin{align*} \lambda\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \right) = \infty = \infty^g = \lambda((x,y)^T)^g \end{align*}
iv) $y = 0, cx + dy = 0, c \ne 0$
Then we must have $x = 0$. Hence $\lambda((0,0)^g) = \lambda((0,0))$ and this should equal $\lambda((0,0))^g$, hence $\lambda((0,0))^g$ must be a fixed point of each $g$ with the above conditions. But I do not see what this fixed point should be? The only reasonable choice would be $\lambda((0,0)) = \infty$, but that does not works out as for $c \ne 0$ we have $\infty^g = a/c$.
v) $y \ne 0, cx + dy = 0$.
\begin{align*} \lambda\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \right) = \infty \end{align*} Here $\lambda((x,y)) = x/y$ and $(x/y)^g = (a(x/y) + b)/(c(x/y) + d)$, and this must equal $\infty$, but it does only if $y = 0$ (as then $\lambda((x,y)) = \infty)$ or $(-d/c)^g = \infty$ is also another value that gives $\infty$, but in general we do not have $x = -d, y = c$ in this situation. So how to go on?
EDIT: With Derek's answer I am able to resolve my issues:
iv) As this gives $x = 0, y = 0$ the argument does not desribe a line, therefore this case is excluded.
v) Here $cx + dy = 0$ implies $\langle (x,y)^T \rangle = \langle (-d, c)^T \rangle$ if $(-d, c) \ne (0,0)$ (note $(-d,c) = (0,0)$ is not possible anyway). Now $$ \lambda(\langle (x,y)^T \rangle^g) = \lambda(\langle (ax + by,0)^T \rangle) = \infty. $$ On the other side we have $\lambda(\langle (x,y)^T \rangle) = \lambda(\langle (-d, c)^T \rangle) = -d/c$ if $c \ne 0$, and $\lambda(\langle (-d, c)^T\rangle) = \infty$ if $c = 0$. An now by convention $(-d/c)^g = \infty$ for $c \ne 0$ and $\infty^g = \infty$ (see wikipedia:Möbius transformation). So that the equality asserted for equivalent actions holds.
Now suppose $c = 0$, then by $ad - bc = 1$ we must have $d \ne 0$ and by convention $\infty^g = \infty$ then. Further as above $\lambda(\langle (x,y)^T \rangle^g) = \infty$ and $\lambda(\langle (x,y)^T \rangle) = \infty = \infty^g$ the equation for equivalent actions hold.
If $d = 0$, then $c \ne 0$ and $b \ne 0$, and $y \ne 0$ together with $cx + dy = 0$ implies $x = 0$. This gives $$ \lambda(\langle (0,y)^T \rangle^g) = \lambda(\langle (by, dy)^T \rangle) = \lambda(\langle (b, 0)^T \rangle) = \infty. $$ On the other side $\lambda(\langle (0,y) \rangle) = 0$, so we must have $0^g = \infty$ if $d = 0$ and the argument ist $0$? I guess this would be natural convention by looking at the mapping side, as then $g$ would be the mapping $$ \frac{ax + b}{cx} $$ and for $x = 0$ setting the value of this "division by zero" to $\infty$ might be a natural definition (anyway, the convention $0^g = \infty$ for $d = 0$ is included in $(-d/c)^g = \infty$).
For all $n>1$, the natural action by matrix multiplication of ${\rm GL}(n,F)$ on $F^n$ induces a (doubly transitive) action on the set $\Omega$ of $1$-dimensional subspaces of $F^n$. This induced action has the scalar matrices in its kernel, so there is an induced action of ${\rm PGL}(n,F)$ on $\Omega$. When $n=2$, this is equivalent to the action on $F \cup \{ \infty \}$ that you are describing.
The equivalence $\lambda$ between the actions is as you defined it, but with $\Omega$ equal to the $1$-dimensional subspaces rather than the vectors of $F^2$. So $\lambda(\langle (x,y)^{\rm T}\rangle)=x/y$ if $y \ne 0$, and $\lambda (\langle (x,0)^{\rm T}\rangle)= \infty$.
All of this is also true with $\rm (P)SL$ in place of $\rm (P)GL$.