I refer to section 8.3, page 119 of Algebra, A Computational Introduction by John Scherk. It is about group action of $GL(2, F)$ on the projective line $P(F) = F \cup \{\infty\}$.
Given a matrix $\alpha = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in GL(2, F)$ for some field $F$, define
$$s_\alpha(x) = \frac{ax + b}{cx + d},\qquad x \in F, x \neq -d/c$$ and $$s_\alpha(-d/c) = \infty,\qquad s_\alpha(\infty) = a/c.$$
The book says it is easy to check that we have defined an action of $GL(2,F)$ on $P(F)$. The problem I'm having is that since $s_\alpha$ is not defined on $\infty \in P(F)$ for any matrix $\alpha$ with $c=0$, we do not have a mapping from every member of $GL(2,F) \times P(F)$ to $P(F)$. So by definition, $s$ is not a group action. Did the book forget to define $s_\alpha(\infty) = \infty$ when $c = 0$, or is there something I'm missing here?
You are right in assuming that iff $c=0$, we define $s_\alpha(\infty) = \infty$. As in the case $c = 0$, we know that $\alpha$ (considered as a linear map on $F^2$) maps $F \times \{0\}$ onto itself. If we consider $P(F)$ as $F^2-\{0\} / \sim$, where $\sim$ is defined by $x \sim y$ iff for some $\lambda \in F^\times$ we have $\lambda x = y$, then $F^\times \times\{0\}$ is the class corresponding to $\infty$. That is $s_\alpha(\infty) = \infty$.