One-to-one function (Linear Algebra) According to [Linear Algebra, 4th, Friedberg, Insel, Spence]
Let $V$ and $W$ be vector spaces and $T : V \rightarrow W$ be linear. 
$(a)$ $T$ is injective if and only if $T$ carries linearly independent subsets of $V$ onto linearly independent subsets of $W$.
$(b)$ Suppose $T$ is injective and $S \subseteq V$. Prove that $S$ is linearly independent if and only if $T(S)$ is linearly independent.
I dont understand the different between these two exercises. Does the first one involve some kind of "surjectivity" since it says map "onto" ? For me the method to prove (a) is let $S$ be linearly independent subset of $V$, then prove $T(S)$ is linearly independent which is certainly (b). 
 A: Let's break down the questions into what you have to prove:
(a)(i) If $T$ is injective, and $S$ is linearly independent, then $T(S)$ is linearly independent.
(a)(ii) If, given any linearly independent $S$, $T(S)$ is linearly independent, then $T$ is injective.
(b)(i) If $T$ is injective, and $S$ is linearly independent, then $T(S)$ is linearly independent.
(b)(ii) If $T$ is injective and $T(S)$ is linearly independent, then $S$ is linearly independent.
As you can see, (a)(i) and (b)(i) are the same, but (b)(ii) does not immediately follow from part (a). It's not quite the same as (a)(ii), as the conclusions and the assumptions are swapped around, in a sense. So no, the question doesn't ask you the same thing, but it's a question that masquerades as a question implicitly with $4$ parts, but really only has $3$.
To prove part (b)(ii) from part (a), you could use the fact that $T$ is invertible on its range. Note that $T(S)$ is a linearly independent subset of the range of $T$, show that the inverse $T'$ is also injective, then use the fact that $S = T'(T(S))$, where $T'$ is injective and $T(S)$ is linearly independent. By (a), $S$ is also linearly independent.
