Let $V$ be a vector space, $N$ a subspace of $V$, and let $X = (w_i)_{i\in I}$ be a family of vectors in $V$ and $Y = \mathrm{span}(X)$.
Prove that $X' = (w_i + N)_i$ is linear independent in the quotient space $V/N$ $\Leftrightarrow$ $X$ is linear independent in $V$ and $N \cap Y = \{0\}$.
So I am struggling to understand what is asked. For the first direction we have an affine subspace in the set of all cosets (which is the set of all equivalence classes of $V$, that are affine subspaces as well?) that is linear independent, which means that $\mathrm{span}(X')= V/N$, right? And from there I have to conclude that $X$ is linear independent.
"$\Rightarrow$":
Consider a linear combination of vectors of $X = (w_i)_{i \in I}$ which is zero. Take the "same" linear combination (same coefficients) but now with the vectors of $X'$. Since the former one was zero so is the latter one. Hence, by assumption, all coefficients are zero.
Let $y \in N \cap Y$. Since $y \in N$ we have that $y + N = N$, which is the zero element in $V / N$. Moreover since $y \in Y$ it can be written as a linear combination of vectors in $X$. Do the same as in the first paragraph (,i.e. make it a linear combination with $X'$). What do you conclude?
"$\Leftarrow$":
Consider an arbitrary linear combination (with vectors of $X'$) which is the zero element in $V / N$. By the definition of the quotient space you obtain that "this" linear combination (with vectors of $X$) is inside of $N$. But this linear combination is clearly inside of $Y$. Hence ... ?