Let [ x ] ∈ M / ∼ {\displaystyle {}[x]\in M/\sim } be given. The only possibility for φ ¯ {\displaystyle {}{\overline {\varphi }}} is to set φ ¯ ( [ x ] ) := φ ( x ) {\displaystyle {}{\overline {\varphi }}([x]):=\varphi (x)} . We have to show that this mapping is well-defined, that is, independent of the choice of the representatives. For this, let [ x ] = [ y ] {\displaystyle {}[x]=[y]} , that is, x ∼ y {\displaystyle {}x\sim y} . By the condition on φ {\displaystyle {}\varphi } , we have φ ( x ) = φ ( y ) {\displaystyle {}\varphi (x)=\varphi (y)} .
![{\displaystyle {}[x]\in M/\sim }](../../../../09e6ad8b0762f878552d79a7992d41aa87c8949d.svg)
be given. The only possibility for 
is to set
![{\displaystyle {}{\overline {\varphi }}([x]):=\varphi (x)}](../../../../0aeb51915bc1025ad46445ad1264769ddafe76c2.svg)
.
We have to show that this mapping is well-defined, that is, independent of the choice of the representatives. For this, let
![{\displaystyle {}[x]=[y]}](../../../../b0e31c58238c766c691394b82eb0fc87a5ee0e3f.svg)
,
that is,
.
By the condition on 
, we have
.