Proof for vector space structure on $V/U $

Question

Task:

Assume $K$ is a field, $V$ is a vector space over $K$ and $U \subset V$ is a vector subspace. Let $V/U$ be the set of the equivalence classes regarding the equivalence relation defined by $U$ ($u \sim v$ iff $u - v \in U$) on $V$ and let $q: V \rightarrow V/U$, $v \rightarrow [v]$ be the canonical projection.

Then there is a unique vector space structure on $V/U$ such that $q$ is a linear map.

Proof:

Since the canonical projection should be linear, the addition must be given by $[v] + [w] = [v + w]$ and the scalar multiplication by $a[v] =[av]$.

Especially there can only be one vector space structure with the required properties, and we have to show that we have well defined operations on $V/U$ that are independently from the representative. This means that for $[v] = [v']$ and $[w] = [w']$ we have to show that $[v + w] = [v' + w']$

Well, now he is showing why this is true for the addition and the scalar multiplication. He then finishes the proof with:

Since the conjunction is well defined and the map is surjective, it follows that we have indeed a vector space structure and that the map is linear.

Questions:

1. I don't understand why he wants to show that $[v] + [w] = [v + w]$. What does this even mean? Sure, we somehow sum up two equivalence classes, but how can I image doing that?

2. Furthermore, it feels like he proves this statement by showing that $[v] = [v']$ and $[w] = [w'] \Rightarrow [v + w] = [v' + w']$. But why?

3. And last but not least, why does it follow that the map is linear? Just because wie map into a vector space, this map doesn't have to be linear, does it?

Answer 1

It might be preferable to work with an example which is easier to get a feeling that abstract symbolic operations.

Take the vector space $V$ to be vectors in a plane. A subspace $U$ to be some line passing through origin (the zero vector) in the plane. (There are no axes, just origin). The cosets $V/U$ are simply the collection of all the lines of the plane parallel to $U$.

Note that these lines cover the whole plane and are disjoint pairwise. To add two cosets $L, L'$ pick one vector from each of these lines. Add these two vectors and find the unique line $L''$ that contains the sum of those two vectors. We declare $L+L'$ to be $L''$: to say this operation is well-defined you must check that different choices of vectors from the same lines even if gives different vector as sum, will be in the same line $L''$.

The linear map $V$ to $V/U$ is given a vector find the line parallel to $U$ containing the given vector (Euclid's fifth postulate). Check it is linear.