# Proof

(related to Proposition: Quotient Space)

• 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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### References

#### Adapted from CC BY-SA 3.0 Sources:

1. Brenner, Prof. Dr. rer. nat., Holger: Various courses at the University of Osnabrück