Compact Hausdorff Spaces: Star-Isomorphic Unital C*-Algebras

[HeinzBor]

Homework Statement

ismorphism between hausdorff compact spaces on C(X) to C(Y)

Relevant Equations

Standard Algebra definitions.

If $X$ and $Y$ are homeomorphic compact Hausdorff spaces, then $C(X)$ and $C(Y)$ are $star$-isomorphic unital $C^{*}$-algebras.

So I got the following map to work with

(AND RECALL THAT $C(X)$ and $C(Y)$ are vector spaces).
$$C(h) : C(Y) \rightarrow C(X) \ : \ f \mapsto f \circ h$$
and thus to show linearity (one of the many statements) I must show that
$$((f+ \lambda g) \circ h)(x) = (f \circ h + \lambda g \circ h) (x) \ \ \lambda \in \mathbb{C}, g \in C(Y)$$
However, I am a bit confused on how to work with this mapping, any help is appreciated..

 

[fresh_42]

Isn't this simply the definition of the vector space property of $C(Y)$?

 

[HeinzBor]

Isn't this simply the definition of the vector space property of $C(Y)$?

Yea I assume I can work with the axioms of a vector space since the functions belong to C(X) and C(Y). However, I haven't had an algebra course for so long, so I am a bit confused on how to work with those definitions, when I have the composition instead of the usual product showing up. Am I allowed to just work with the composition as a standard product in a vector space?

 

[fresh_42]

You simply have $h(x)=:y\in Y$. So if you write down the linearity in $C(Y),$ namely the linearity
$$
(f+\lambda \cdot g)(y) := f(y)+ \lambda \cdot g(y),
$$
in a function space, then you get exactly what you need. Only substitute $y$ by $h(x).$

 

[HeinzBor]

You simply have $h(x)=:y\in Y$. So if you write down the linearity in $C(Y),$ namely the linearity
$$
(f+\lambda \cdot g)(y) := f(y)+ \lambda \cdot g(y),
$$
in a function space, then you get exactly what you need. Only substitute $y$ by $h(x).$

Ahhh yes ! of course.. $$X \rightarrow V$$ can be given the structure of a vector space over $$\mathbb{F}$$, if for any $$f,g : X \rightarrow V$$, $$x \in X$$, $$\lambda \in \mathbb{F}$$ $$(f+g)(x) = f(x) + g(x)$$ and $$(\lambda f)(x) = \lambda f(x)$$ and composition is associative..

 

[HeinzBor]

Okay so for linearity:

$$
((f + \lambda g) \circ h)(x) = f((h(x)) + \lambda g (h(x)))
= (f \circ h + \lambda g \circ h)(x)
$$

Multiplicativity:

$$
(f \circ g) ( g \circ h) (x) = f(g \circ h)(x)
= ((fg) \circ h) (x).
$$

To show the mapping preserves involution:

$$
(\overline{f} \circ h)(x) = \overline{f}(x)
= (\overline{f \circ h})(x)
$$Lastly I must show that it is unital i.e.

$$1 h = 1$$, somehow it seems really obvious, but how can h just be gone... $$h$$ was assumed to be a homeomorphism between to compact hausdorff spaces so its 1-1 and has continuous inverse...