$w_1, \dots , w_n \in W$ are linearly independent, prove that so also $v_1, \dots, v_n \in V$, if $f(v_i) = w_i$ 
Let $f: V \rightarrow W$ be a linear map. If $w_1, \dots , w_n \in W$ are linearly independent, prove that so also $v_1, ..., v_n \in V$, if $f(v_i) = w_i$, for $i = 1,\dots,n$.

I have tried to figure this out, but the problem is that I don't know the definition of $f$, so maybe it could make vectors that are linearly dependent $v_i$ to vectors that are linearly independent $w_i$.
 A: Outline: suppose as usual you have scalars $\lambda_1,\dots,\lambda_n$ such that 
$$
0_V = \sum_{k=1}^n \lambda_k v_k.
$$
The goal is to prove that one must have $\lambda_1=\dots=\lambda_n = 0$. To do so, apply $f$ to both sides:
$$
0_W=f(0_V)= f\left( \sum_{k=1}^n\lambda_k v_k \right) = \sum_{k=1}^n\lambda_k f(v_k) = \sum_{k=1}^n\lambda_k w_k
$$
where we used the fact that $f$ was a linear map to "take out" the sum and the scalars. Knowing that the $w_k$ are independent, can you conclude?
A: Prove the contrapositive: If the $v_1,\ldots,v_n$ are linearly dependent, then so are the $w_1,\ldots,w_n$.
So assume
$$a_1v_1+\ldots+a_nv_n=0$$
Then
$$a_1w_1+\ldots+a_nw_n=a_1f(v_1)+\ldots+a_nf(v_n)$$
$$=f(a_1v_1+\ldots+a_nv_n)$$
$$=f(0)=0$$
Two steps there depend on the linearity of $f$.
A: Take a linear combination
$$\sum_{i=1}^n\lambda_i v_i=0$$
Apply $f$ to this equality
$$\sum_{I=1}^n\lambda_i f(v_i)=\sum_{i=1}^n\lambda_i w_i=0$$
The $(w_i)_i$ being independent allows to conclude $\forall i\,\lambda_i=0$ and therefore the $(v_i)_i$ are linearly independent
