In page 2 of the book Differential geometry: Cartan's generalization of Klein's Erlangen program by Sharpe, it stated the example of the projective plane $P(\mathbb{R}^{n+1})$.
The book said if $a:\mathbb{R}^n\rightarrow \mathbb{R}^{n+1}$ is an affine map whose image does not contain the origin, then $p \circ a$ is injective and continuous, where $p:\mathbb{R}^{n+1} \setminus\{0\} \rightarrow P(\mathbb{R}^{n+1})$ is the quotient map.
I am confused with this part, since what I know about affine map is a map in the form $a(x)=Ax+b$, where $A: \mathbb{R}^{n} \rightarrow\mathbb{R}^{n+1}$ is a linear transformation and $b \in \mathbb{R}^{n+1}$. In particular, if we take $A$ to be a zero transformation, then it is impossible for the map $p \circ a$ to be injective.
Is that an error in the book or else?
You're right that to have any hope of $p \circ a$ being injective, we need $a$ to be injective; in particular, if $a(x) = Ax + b$, then the matrix $A$ must have full column rank. (I don't have a copy of the book, so I can't say if this is an error or merely an assumption you've missed.)
The point of the theorem is that the quotient map $p$ takes lines through the origin to points in projective space, and the image of any affine map usually intersects such a line in at most one point. The exception is when the image contains the origin; in that case, except in the very degenerate case when the image is $0$-dimensional, there will be some line(s) through the origin entirely contained in the image of $a$.
As a result, when the image of $a$ doesn't contain $0$, there can never be any "further loss of injectivity" by applying $p$.