Question #162936

Prove that if phi : S ---> T is an isomorphism of <S,*> with <T, #> and psi: T ---> U is an isomorphism of <T, #> with <U, diamond> then the composite function psi composed with phi is an isomorphism of <S, *> with <U, diamond>


1
Expert's answer
2021-02-19T15:18:41-0500

To show that ψϕ\psi \circ \phi is an isomorphism, we show that ψϕ\psi \circ \phi is a homomorphism and then we show that it is bijective.

To show homomorphism: Let x,ySx,y \in S consider,

ψϕ(xy)=ψ(ϕ(xy))\psi \circ \phi (xy) = \psi (\phi (xy)) . Since ϕ\phi is a homomorphism, we have that

ψ(ϕ(xy))=ψ(ϕ(x)ϕ(y))\psi (\phi (xy)) = \psi (\phi(x) \phi(y))

And since ψ\psi is a homomorphism, we have that

ψ(ϕ(x)ϕ(y))=ψ(ϕ(x))ψ(ϕ(y))=ψϕ(x)ψϕ(y)\psi (\phi(x) \phi(y)) = \psi(\phi(x))\psi(\phi(y)) = \psi \circ \phi(x)\psi \circ \phi(y)

as desired. Hence ψϕ\psi \circ \phi is a homomorphism.

To show that it is bijective, we show that ψϕ\psi \circ \phi is both injective and surjective.

Let ψϕ(x)=ψϕ(y)\psi \circ \phi (x) = \psi \circ \phi (y) for x,ySx,y \in S

    ψ(ϕ(x))=ψ(ϕ(y))\implies \psi (\phi(x)) = \psi (\phi(y))

And since ϕ\phi is injective, we have that

ψ(x)=ψ(y)\psi(x) = \psi(y)

And since ψ\psi is injective, we have that

x=yx=y .

Let zUz \in U , then pTψ(p)=z\exist p \in T \ni \psi(p) = z

(Since ψ\psi is surjective).

Since pTp \in T , by the surjectivity of ϕ\phi, we have that qSϕ(q)=p\exist q \in S \ni \phi(q) =p

So we have that z=ψ(p)=ψ(ϕ(q))=ψϕ(q)z = \psi(p) = \psi(\phi(q)) = \psi \circ \phi(q)

And this implies that ψϕ\psi \circ \phi is surjective. Hence we can safely conclude that ψϕ\psi \circ \phi is an isomorphism.


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