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Let $V$ be a Vector Space and $N$ a subspace of $V$, and let $X = (w_i)_{i\in I}$ be a family of vectors in $V$ and $Y = span(X)$.

Proof 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 set of all equivalence classes of $V$, that are affin subspaces as well?) that is linear independent, that means that $span(X')= V/N$, right? And from there I have to conclude that $X$ is linear indpendent.

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"$\Rightarrow$":

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 ... ?

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    Why is N the zero Element in V/N? – 2017-01-10
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    ... by definition: *Elements* in $V/N$ are cosets/affine spaces "of" $N$, i.e. are of the form $v + N$, where $v \in V$. Look at the definition of the *addition* in this space and you will see that $0 + N = N$ is the zero element w.r.t this operation/addition. – 2017-01-10
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    Thanks, but why is in "<=" suddenly X' the zero element? are the linear combinations of X' = N? – 2017-01-12
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    I meant that you should consider a linear combination with vectors from your set $X' = (w_i + N)_i$. And this linear combination is zero, i.e. $\sum_i a_i \cdot (w_i + N) = 0_{V / N} = N$. – 2017-01-12