How to prove the unique linear map that extends a function is only injective iff the vectors of the function are linearly independent Let $F$ be a field, let $V_1,V_2$ be vector spaces over F, and suppose $B = ({u_1,u_2,...,u_k})$is a basis of $V_1$ (where $k=dimV_1$).Consider functions $φ,ψ:B→V,$
and write $f$ and $g$ for the linear extensions from $V_1$ to $V_2$ of $φ$ and $ψ$ respectively.
Prove f is injective iff the vectors  $φ(u_1).... φ(u_k)$ are linearly independent.
Attempt:
Suppose the set $\{φ(v_{1}),φ(v_{2}),\ldots,φ(v_{n})\}\subset V_{1}$ is linear independent and the linear mapping $f:V_{1}\rightarrow V_{2}$ in injective. Now
\begin{align*}
& \alpha_{1}f(φ(v_{1})) + \alpha_{2}f(φ(v_{2})) + \ldots + \alpha_{n}f(φ(v_{n})) = 0 \Longrightarrow f(\alpha_{1}φ(v_{1}) + \alpha_{2}φ(v_{2}) + \ldots + \alpha_{n}φ(v_{n})) = 0\Longrightarrow\\\\
& \alpha_{1}φ(v_{1}) + \alpha_{2}φ(v_{2}) + \ldots + \alpha_{n}φ(v_{n}) = 0 \Longrightarrow \alpha_{1} = \alpha_{2} = \ldots = \alpha_{n} = 0
\end{align*}
Thus the set $\{f(φ(v_{1})),f(φ(v_{2})),\ldots,f(φ(v_{n}))\}$ is linear independent.
Now I'm stuck on this second part.
$g$ is surjective if and only if the set ${ψ(u) : u ∈ B}$ is a spanning set of $V_2$.
So I know for this one we know that since $B$ is a basis, then B is a spanning set of $V_1$ but I'm unsure how to continue? Can anyone check my proof and assist with this one?
 A: $\textbf{For the first part:}$
$\phi(u_1),...,\phi(u_k)$ independent $\implies f$ injective:
if $f(a) = f(b)$, then
$f(\sum_{i=1}^k c_i u_i) = f(\sum_{i=1}^k c_i' u_i)$
$\iff \sum_{i=1}^k c_i f(u_i) = \sum_{i=1}^k c_i' f(u_i)$
$\iff \sum_{i=1}^k (c_i - c_i') f(u_i) = 0$
$\iff \sum_{i=1}^k (c_i - c_i') \phi(u_i) = 0$ 
$\implies c_i-c_i'=0$ $\forall i$, since $\phi(u_1),...,\phi(u_k)$ independent
For the converse, if $\phi(u_1),...,\phi(u_k)$ dependent, then $\exists c_1,...c_k$ not all $0$ such that 
$\sum_{i=1}^k c_i \phi(u_i) = 0$
$\iff f(\sum_{i=1}^k c_i u_i) = 0$
but $f(0) = 0$ since $f$ is linear 
and $\sum_{i=1}^k c_i u_i \ne 0$ since not all $c_i$ are $0$ and $u_i$ are a basis
so $f$ is not injective
$\textbf{For the second part:}$
$\psi(u_1),...,\psi(u_k)$ span $\implies g$ surjective:
Let $v \in V_2$, then since $\psi(u_1),...,\psi(u_k)$ span, $\exists c_1,...,c_k$ such that 
$v = \sum_{i=1}^k c_i \psi(u_i)$
$= \sum_{i=1}^k c_i g(u_i)$
$= g(\sum_{i=1}^k c_i u_i)$
For the converse, just note for $v \in V_2$,
$v = g(a) = g(\sum_{i=1}^k c_i u_i) = \sum_{i=1}^k c_i g(u_i) = \sum_{i=1}^k c_i \psi(u_i)$
