Proof

By hypothesis, \(V\) is a vector space over a field \(F\), and \(U\subseteq V\) is its subspace defining the equivalence relation induced by \(U\) on \(V\) with the quotient set \(V/U\).
Because the canonical projection \(q\colon V\longrightarrow V/U,\,v\longmapsto [v]\,,\) must become a linear map over the field \(F\), it has to fulfill the properties
- addition \([v]+[w]=[v+w]\) for all equivalent classes in \(V/U\) and
- scalar multiplication \(\lambda [v]=[\lambda v]\) for all \[[v]\in V/U\] and \(\lambda \in F\).
Because the above properties have to be valid for all elements, there can only be only one quotient space \(V/U\) such that \(q\) is a linear map over \(F\).
The above properties do not depend on the representatives of the equivalence classes.
- Let \([v]=[v']\) and \([w]=[w']\). We have to show that \([v+w]=[v'+w']\). By hyothesis, we can set \(v'=v+u\) and \(w'=w+u'\) with some \(u,u'\in U\). It follows \[v'+w'=v+w+u+u'\,\] and the representative \(v+w+u+u'\) is equivalent to the representative \(v+w\) because of \(u+u'\in U\).
- Similarly, let \(v'=v+u\) with \(u\in U\). Then it follows \[\lambda v'=\lambda (v+u)=\lambda v+\lambda u\,,\] and the representative \(\lambda v+\lambda u'\) is equivalent to \(\lambda v\) because of \(\lambda u \in U\).
Because the addition and the scalar multiplication on \(V/U\) are well-defined and because \(q\) is surjective it follows that \(V/U\) \(q\) is linear map and that the quotient space \(V/U\) is a vector space.
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Brenner, Prof. Dr. rer. nat., Holger: Various courses at the University of Osnabrück