When is a function one-to-one? I am going crazy over a statement that is supposed to be false.
"A function T: R^n -> R^m is one-to-one if the only vector v in R^n whose image is 0 is v = 0"
I thought this was true since otherwise if there was another vector u that did not equal the zero vector but whose image was the zero vector then f(u) and f(v) would not be distinct and hence the function not one-to-one. This also explains why the null space consisting of only the zero vector makes the function one-to-one. For if in the RREF there were free variable(s) then the generating set for the null space would not only be the zero vector. Multiple vectors of R^n would equal the zero vector of R^m and hence the function would not be one-to-one.
Where does my logic break?
 A: The definition of being one to one is: if $f(x)=f(y)$ then $x=y$.
This definition is for functions. The statement you are posting don't work for functions in general, it's works for a special kind of functions called linear (maybe linear transformation in your case). Then
If $T$ is just a function: this is false. Take $T:\Bbb{R}\to\Bbb{R}$ such that $T(0)=0$ and $T(x)=1$ if $x\neq 0$.
If $T$ is linear: this is true. Suppose that the unique vector $\mathbf{v}$ such that $T(\mathbf{v})=0$ is $\mathbf{v}=0$. If $T(\mathbf{v})=T(\mathbf{w})$ then $T(\mathbf{v})-T(\mathbf{w})=0$, so $T(\mathbf{v-w})=0$ which implies $\mathbf{v}=\mathbf{w}$ and $T$ is one to one. The other direction is clear.
A: Now it depends on the definition of one-to-one. For me one-to-one means that
the mapping is injective and surjective. So $T: R \to R^2, x\mapsto (x,0)$ would not be one-to-one in this definition.
Since it is said in a comment that this does not answer the question: The question is why a -- I guess linear -- mapping from $R^m$ to $R^n$ can be not one-to-one if only the zero vector is mapped into the zero vector. If an one-to-one correspondence is meant as I suppose here, it gives an example and answers the question.
