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If X is a locally compact Hausdorff space and $X=X_1 \cup X_2$ where $X_1, X_2$ are disjoint open and closed subsets X, I want to show that $C_0(X)$ isomorphic to $C_0(X_1) \oplus C_0(X_2)$

I have that $K_0(A \oplus B) \cong K_0(A) \oplus K_o(B)$ but I am having some trouble seeing exactly how this helps me with $C_0$ as I am quite new in this area but this is what's "closest" to what I want to end up with. Is there a nice way to say that :

$$C_0(X) = C_0(X_1 \cup X_2 ) \cong C_0(X_1 \oplus X_2 ) $$

And is this even true? Also if $K_0(X) \cong C_0(X)$ somehow, such that it simply follows from my proposition in the book, this would be nice, but I can't seem to find an answer. Or is there another approach to this type of problem?

Miep
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  • Forgetting about K-theory entirely, this seems easy to solve by hand: first write down the obvious map from $C_0(X)$ to $C_0(X_1) \oplus C_0(X_2)$ and then check that it is an isomorphism. (Depending on exactly what kind of isomorphism you are asking about - isomorphism of rings? Banach spaces? $C^*$-algebras? - the exact details of what you have to prove may vary slightly.) – Nate Eldredge May 04 '20 at 18:15
  • I assume $C_0(X)$ is the space of continuous functions on $X$ vanishing at infinity? – Nate Eldredge May 04 '20 at 18:19
  • @NateEldredge Yes $C_0(X)$ is exactly that and it should be an isomorphism of $C^*$-algebras – Miep May 04 '20 at 18:27
  • Okay, so then you have to show that the obvious map is a *-algebra homomorphism, an isometry, and surjective. Is there a particular one of these steps where you get stuck? – Nate Eldredge May 04 '20 at 18:28
  • This is probably a stupid question but I just need to make sure on the obvious map and how to formulate it properly. So the obvious map would be the one that takes f(a) to $f(a,0)$ if $a \in X_1$ and to $f(0,a)$ if $f \in X_2$ right? – Miep May 04 '20 at 18:57
  • Well, I don't see how what you've stated is taking values in $C_0(X_1) \oplus C_0(X_2)$. Remember that the latter is typically defined as the set of ordered pairs $(f_1, f_2)$ with $f_i \in C_0(X_i)$. So to me, the "obvious map" would be $f \mapsto (f|{X_1}, f|{X_2})$, where $f|_A$ denotes restriction. – Nate Eldredge May 04 '20 at 18:59
  • Arh, yes this makes more sense. Thank you! I'll give it a try :-) – Miep May 04 '20 at 19:04

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