Equivalence relation/Quotient set/Universal property/Fact/Proof
Proof
Let be given. The only possibility for is to set . We have to show that this mapping is well-defined, that is, independent of the choice of the representatives. For this, let , that is, . By the condition on , we have .
![{\displaystyle {}[x]\in M/\sim }](https://wikimedia.org/api/rest_v1/media/math/render/svg/09e6ad8b0762f878552d79a7992d41aa87c8949d)
![{\displaystyle {}{\overline {\varphi }}([x]):=\varphi (x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0aeb51915bc1025ad46445ad1264769ddafe76c2)
![{\displaystyle {}[x]=[y]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0e31c58238c766c691394b82eb0fc87a5ee0e3f)
