- #1
Bachelier
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WTS is that [itex]\mathbb R^*/N \ \cong \ \mathbb R^{**}[/itex] where [itex]N = (-1, 1)[/itex]
then prove that [itex]\mathbb R^*/\mathbb R^{**} \ is \ \cong \ to \mathbb Z/2\mathbb Z[/itex]
So the best answer in my opinion is to construct a surjection and use the first iso thm.
[itex]f:\mathbb R^*\rightarrow\mathbb R^{**}[/itex]
[tex]f(x)=|x|,[/tex] is onto by construction. clearly a homomorphism
[itex]Ker \ (f) = N[/itex], hence [itex]\mathbb R^*/N \ \cong \ \mathbb R^{**}[/itex]
part 2
[itex]ψ:\mathbb R^*\rightarrow\ \ N[/itex]
[tex]ψ(x)=1 \ if \ x>0 \ and \ ψ(x)=-1 \ if \ x<0[/tex]
by same thm, [itex]\mathbb R^*/\mathbb R^{**} \cong \mathbb Z/2\mathbb Z[/itex]
because it has 2 elements one of each is the identity.
then prove that [itex]\mathbb R^*/\mathbb R^{**} \ is \ \cong \ to \mathbb Z/2\mathbb Z[/itex]
So the best answer in my opinion is to construct a surjection and use the first iso thm.
[itex]f:\mathbb R^*\rightarrow\mathbb R^{**}[/itex]
[tex]f(x)=|x|,[/tex] is onto by construction. clearly a homomorphism
[itex]Ker \ (f) = N[/itex], hence [itex]\mathbb R^*/N \ \cong \ \mathbb R^{**}[/itex]
part 2
[itex]ψ:\mathbb R^*\rightarrow\ \ N[/itex]
[tex]ψ(x)=1 \ if \ x>0 \ and \ ψ(x)=-1 \ if \ x<0[/tex]
by same thm, [itex]\mathbb R^*/\mathbb R^{**} \cong \mathbb Z/2\mathbb Z[/itex]
because it has 2 elements one of each is the identity.