I have tried multiple approaches to this following problem, but none have been successful so far. I just don't know where to begin, does anyone have some tips?
Suppose $f:V\to V$ is a linear mapping. Define $S=\ker(f), W=V/S$ and $f':W\to W$ the induced mapping. Regard the following statements:
(a) $v_1,v_2...v_n$ are independent in $V$
(b) $v_1+S,v_2+S,...v_n+S$ are independent in W
Prove or disprove $(a)\implies(b)$ and $(b)\implies(a)$.
First I looked at what happened if you took $f(\lambda_1v_1+...+\lambda_nv_n)=0$ and $f'(\lambda_1(v_1+S)+...+\lambda_n(v_n+S))$ and things alike but it didn't really get me anywhere.
I also looked at a basis for $V$ consisting of $x_1,...,x_j, y_1,...,y_k$, a basis for $W$ consisting of $x_1+S,...,x_j+S$ and a basis for $S$ consisting of $y_1,...,y_k$. With $v_1,...,v_n$ being independent and in $V$, we know $$v_1,...,v_n\in \{x_1,...,x_j,y_1,...y_k\} \implies $$ $$v_1+S,...,v_n+S\in\{x_1+S,...,x_j+S,y_1+S,...,y_k+S\}$$ but $\{x_1+S,...,x_j+S\}$ is a basis for $W$, so $x_1+S,...,x_j+S,y_1+S,...,y_k+S$ are not independent because it adds to the basis. Therefore, $v_1+S,...,v_n+S$ do not have to be linearly independent.
That's all I had so far, and I suspect that $(b)\implies(a)$ might be true with a relatively easy proof. I have yet to understand that though.
I would disprove $(a) \Rightarrow (b)$ like this: take $f\equiv0$, then $S=V$, $W\cong 0$ and for any $v_1,\ldots,v_n$ independent holds $v_1+S=\ldots =v_n+S$, so the statement is not true for (this $f$ and) any $V\ncong 0$ (why?), so it doesn't hold in general.
For the other direction: Assume $v_1,\ldots,v_n$ are dependent, e.g. $\mu_1 v_1+\ldots +\mu_n v_n = 0$ with $\mu_i$ not all equal to $0$. Let g denote the canonical injection of $V$ into $W$.$^1$ We have $$g(\mu_1 v_1+\ldots \mu_n v_n) = \mu_1 g(v_1)+\ldots +\mu_n g(v_n) = \mu_1(v_1+S)+\ldots+\mu_n(v_n+S).$$
Since $g(0)=0$ and our $\mu_i$ haven't changed, $v_1+S,\ldots,v_n+S$ are also dependent. Proof by contraposition.
$^1$ The canocical injection of $V$ into $W$ is defined as such: Let $(s_1,\ldots,s_k)$ be a basis of $S=\ker f$ and let $(s_1,\ldots,s_k,t_1,\ldots,t_\ell)$ be a basis of $V$. For any $v=\lambda_1 s_1+\ldots+\lambda_k s_k+\nu_1 t_1+\ldots+\nu_\ell t_\ell$ set $g(v)=\nu_1 t_1+\ldots+\nu_\ell t_\ell + S$. Since $\lambda_1 s_1+\ldots+\lambda_k s_k\in S$ we have furthermore $$g(v) = (\nu_1 t_1+\ldots+\nu_\ell t_\ell)+(\lambda_1 s_1+\ldots+\lambda_k s_k) + S = v+S$$ $g$ is clearly a linear map, hence $g(0)=0+S$ (or $g(0)=0$ for short).