I'm reading Herstein's Topics in Algebra and Halmos's Finite Dimensional Vector Spaces. I think I understand that if $V_F$ is a vector space over $F$ and $W$ a subspace of $V$, then
- there is an equivalence relation for $u$ and $v$ in $V$ defined by $u \equiv v \pmod W$ when $(u - v) \in W$
- the equivalence classes of the relation correspond to the elements of the quotient space $V/W$.
- therefore a quotient space is always a partitioning of the space into equivalence classes (of the corresponding modulus relation).
I haven't seen any reference to the converse, and this is my main question: Does an equivalence relation always have a corresponding quotient space structure ? I suspect not, and I think that the following is a valid counter-example:
Define the relation $\leftrightarrow$ between $u$ and $v$ by $u \leftrightarrow v$ if $u = \lambda v$ for some non-zero $\lambda \in F$. This appears to be an equivalence relation where the $0$ vector is in a class of its own, and "co-linear" vectors are in corresponding equivalence classes. I can't see any way (of course, that isn't a proof) to set this up as a modulus relationship, and therefore conclude that it isn't represented by a quotient space. So, a secondary question is: Is this a valid equivalence relation, and is it representable as a quotient space - if not, what's the proof ?
(Depending on feedback, I may have a further related question about tensor products)