$T$ is surjective if and only if $f_1,f_2,f_3$ is linearly independent. I was solving exercises and I found the following problem: Suppose you have $V$ an $\mathbb{R}$-vector space (with finite dimension) and $f_1,f_2,f_3$ belonging to the dual space of $V$ $(V^*)$. Let $T$ be the linear transformation
$$T: V \rightarrow  \mathbb{R}^3$$ given by
$$ T(v ) = (f_1(v),f_2(v),f_3(v)) $$ ($v$ belonging to $V$) Prove that $T$ is surjective if and only if ${f_1,f_2,f_3}$ are linearly independent as vectors in $ V^* $.

My first attempt was to find a relation between being surjective (ephimorphism) and injective (monomorphism),
since I remember that they are equivalent if a condition related to the dimension is met, but I am not sure if this is the correct way.
On the other hand, I was looking to write $f_1,f_2,f_3$ as a linear combination and try to show that it is equal to $0$ only if the coefficients are $0$, but I can't see how the surjectivity of $T$ can help me prove it.
$$  af_1(v) + bf_2(v) + cf_3(v) = 0 $$
In general, I can see that if $f_1,f_2,f_3$ were linearly dependent, they couldn't generate all of $\mathbb{R}^3$ so $T$ wouldn't be surjective, but I can't find the right way to prove it.
I'd appreciate some hints.
 A: If $  a.f_1(v) + b.f_2(v) + c.f_3(v) = 0 $ with at least one of $a,b,c$ not zero then $T$ cannot have $a,b,c$ in  its range. This is bacause there would then be a $v$ such that $f_1(v)=1,f_2(v)=b$ and $f_3(v)=c$ which gives $a^{2}+b^{2}+c^{2}=0$ which implies $a=b=c=0$.
Conversely, suppose $f_i$'s are independent. If $T$ is not surjective, then its range would have dimension less than $3$. This implies that there is a non-zero vector $(a,b,c)$ orthogonal to the range. This gives $  a.f_1(v) + b.f_2(v) + c.f_3(v) = 0 $ for all $v$ contradicting linear independence.
A: A linear map between finite-dimensional vector spaces is surjective (resp. injective) if and only if its dual map is injective (resp. surjective).
Hence, $T \colon V \to \Bbb R^3$ is surjective if and only if $T^* \colon (\Bbb R^3)^* \to V^*$ is injective. Now, as $(\Bbb R^3)^*$ is isomorphic to $\Bbb R^3$ via the map
\begin{align}
\Bbb R^3 & \stackrel \varphi \longrightarrow (\Bbb R^3)^* \\
(a_1,a_2,a_3) & \longmapsto \big( (x_1,x_2,x_3) \mapsto a_1x_1+a_2x_2+a_3x_3 \big),
\end{align}
then $T^*$ is injective if and only if $T^* \circ \varphi$ is injective, that is, the map
\begin{align}
\Bbb R^3 & \longrightarrow V^* \\
(a_1,a_2,a_3) & \longmapsto a_1f_1+a_2f_2+a_3f_3,
\end{align}
is injective.
Clearly, this is the same thing as saying that $f_1,f_2$ and $f_3$ are linearly independent.
Note: This solution works even if we replace $\Bbb R$ with a generic field.
