Question about affine transformation and one-one function. An Affine transformation is a function $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ such that $f(v) = Av+b $, where $\det A \neq 0 $ and $b \in \mathbb{R}^2.$ 
My professor give me a task, I need to prove that a function $f$ is Affine transformation iff $f$ is one-one.
It's a little confuse, I have proved that if $f$ is affine then $f$ is one - one, but the other hand I don't know and I suspect it's not true. Of course, if I take $f: \mathbb{R}_{+} \rightarrow \mathbb{R},$ where $f(x) = x^2,$ then this function is one- one and not affine, but I need some example in $\mathbb{R}^2$ with the form $f(v)=Av + b.$

I think you have mis-interpreted the question. What is true is if $f(v)=Av+b$ for some square matrix $A$ and some vector $b$ then $f$ is one-to-one iff $\det(A) \neq 0$. 
Proof: If $\det(A) \neq 0$ and $Av+b=Aw+b$ then $A(v-w)=0$ and this implies $v=w$ because $A$ is non-singular (and its kernel is $\{0\}$). 
If $\det(A)=0$ the there exists a nonzero vecto $v$ such that $Av=0$. This gives $f(v)=f(2v)$ so $f$ is not one-to-one. 
