A textbook example asks you to prove that for the linear transformation $T\colon V \to W$ that it is one-to one if and only if $T$ carries linearly independent subsets of $V$ onto linearly independent subsets of $W$.
The answer key gives the following solution:
Suppose $T$ is one-to-one. Then let $S$ be a linearly independent subset of $V$, and let $S' = \{T(s)\mid s \in S\}$. (Notice that we’re not assuming that $S$ is finite.) Suppose there are some elements of $S’$ $T(s_1), T(s_2), \ldots , T(s_n)$ and scalars $a_1, \ldots, a_n$ such that $a_1T(s_1)+ \cdots + a_nT(s_n) = 0$. Then by the linearity of $T$,
$a_1T(s_1) + . . . + a_nT(s_n) = 0$
$T(a_1s_1) + . . . + T(a_ns_n) = 0$
$T(a_1s_1 + . . . + a_ns_n) = 0$
But $T$ is one-to-one, so $a_1s_1 + \cdots + a_ns_n = 0$, and thus $a_1 = \cdots = a_n = 0$ because $S$ is linearly independent. Thus $S’$ is linearly independent.
I don't understand this solution. Why does $(a_1s_1+...a_ns_n)$ need to equal 0 just because $T(\text{stuff}) = 0$? A one-to-one function doesn't mean that $T(1) = 1$ and $T(2) = 2$ etc... It simply means all numbers in the codomain must appear in the range.